The standard oxidation potential of $Ni/Ni^{ 2+ }(Ni^{ 2+ }=1M)$ electrode is 0.236 V. If this is combined with a hydrogen electrode ($P_{ { H }_{ 2 } }$ = 1 atm ) in acid solution, the pH at which the measured e.m.f of the solution will be zero at ${ 25 }^{ 0 }$C is?
Answer
253.5k+ views
Hint: Here you should know that the standard reduction potential of hydrogen electrodes is always zero. The Nernst equation will be used here and finally you need to calculate pH by taking the negative log. Now you can easily answer this question.
Complete step by step answer:
Here we are provided with a standard oxidation potential of nickel electrode and we all know that the reduction potential of hydrogen electrode is always zero so first the standard electrode potential of the cell is equal to the oxidation potential of the nickel electrode.
$Ni\quad \rightarrow \quad Ni^{ 2+ }\quad +\quad 2e⁻$ ; Oxidation Potential of Electrode $E^{ 0 }_{ OP }$= 0.236V
$2H^{ + }\quad +\quad 2e⁻\quad \rightarrow\quad H_{ 2 }$ ; Reduction Potential of Electrode $E^{ 0 }_{ RP }$ = 0
So we can write, $E^{ 0 }_{ cell }$ = $E^{ 0 }_{ OP }$ + $E^{ 0 }_{ RP }$ = 0.236 + 0 = 0.236 V
Here we are also given that e.m.f of the cell is zero and $Ni^{ 2+ }$ ion concentration is also 1M. So, substitute the values of different terms in Nernst equation and calculate the concentration of hydrogen ions.
$E_{ cell }\quad =\quad E_{ cell }^{ 0 }\quad +\quad \dfrac { 0.0592 }{ 2 } log_{ 10 }[H^{ + }]^{ 2 }-log_{ 10 }[Ni^{ 2+ }]$
$0\quad =\quad 0.236\quad +\quad \dfrac { 0.0592 }{ 2 } log_{ 10 }[H^{ + }]^{ 2 }$
=> $log_{ 10 }[H^{ + }]^{ 2 }$ = -8
=> 2$log_{ 10 }[H^{ + }]$ = -8
=> $log_{ 10 }[H^{ + }]$ = -4
From the concentration one can easily find out the pH as using the following expression that we already know pH = -$log_{ 10 }[H^{ + }]$
So, pH = 4
Therefore, the correct answer to this question is pH = 4.
Note: During calculation of standard cell potential beware about either standard oxidation potential is given or standard reduction potential is given, because reduction potential of metal will always be negative while standard oxidation potential is positive.
Complete step by step answer:
Here we are provided with a standard oxidation potential of nickel electrode and we all know that the reduction potential of hydrogen electrode is always zero so first the standard electrode potential of the cell is equal to the oxidation potential of the nickel electrode.
$Ni\quad \rightarrow \quad Ni^{ 2+ }\quad +\quad 2e⁻$ ; Oxidation Potential of Electrode $E^{ 0 }_{ OP }$= 0.236V
$2H^{ + }\quad +\quad 2e⁻\quad \rightarrow\quad H_{ 2 }$ ; Reduction Potential of Electrode $E^{ 0 }_{ RP }$ = 0
So we can write, $E^{ 0 }_{ cell }$ = $E^{ 0 }_{ OP }$ + $E^{ 0 }_{ RP }$ = 0.236 + 0 = 0.236 V
Here we are also given that e.m.f of the cell is zero and $Ni^{ 2+ }$ ion concentration is also 1M. So, substitute the values of different terms in Nernst equation and calculate the concentration of hydrogen ions.
$E_{ cell }\quad =\quad E_{ cell }^{ 0 }\quad +\quad \dfrac { 0.0592 }{ 2 } log_{ 10 }[H^{ + }]^{ 2 }-log_{ 10 }[Ni^{ 2+ }]$
$0\quad =\quad 0.236\quad +\quad \dfrac { 0.0592 }{ 2 } log_{ 10 }[H^{ + }]^{ 2 }$
=> $log_{ 10 }[H^{ + }]^{ 2 }$ = -8
=> 2$log_{ 10 }[H^{ + }]$ = -8
=> $log_{ 10 }[H^{ + }]$ = -4
From the concentration one can easily find out the pH as using the following expression that we already know pH = -$log_{ 10 }[H^{ + }]$
So, pH = 4
Therefore, the correct answer to this question is pH = 4.
Note: During calculation of standard cell potential beware about either standard oxidation potential is given or standard reduction potential is given, because reduction potential of metal will always be negative while standard oxidation potential is positive.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

Classification of Drugs in Chemistry: Types, Examples & Exam Guide

Types of Solutions in Chemistry: Explained Simply

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

[Awaiting the three content sources: Ask AI Response, Competitor 1 Content, and Competitor 2 Content. Please provide those to continue with the analysis and optimization.]

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Other Pages
CBSE Class 12 Chemistry Question Paper 2026 PDF Download (All Sets) with Answer Key

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

NCERT Solutions For Class 12 Chemistry Chapter 10 Biomolecules - 2025-26

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 12 Chemistry Chapter 2 Electrochemistry - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions - 2025-26

