Answer
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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.
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