Question

If $E_{A{u^ + }/Au}^0$ is $1.69{\text{V}}$ and $E_{A{u^{3 + }}/Au}^0$ is $1.40{\text{V}}$ then $E_{A{u^{3 + }}/A{u^ + }}^0$ will be
A. $0.19{\text{V}}$
B. $2.945{\text{V}}$
C. $1.255{\text{V}}$
D. None of these

Answer 1

Hint: To answer this question you must recall the concept of electrode potentials. The electrode potentials of metals help us to predict the ease of the metal to undergo reduction or oxidation. Using this information, we can select a suitable metal for the anode and cathode electrons. We know that anode is the site for oxidation and cathode for reduction. Accordingly we can choose the suitable electrodes to satisfy the requirements of reactions.

Complete step by step solution:
- The branch of chemistry that deals mainly with the relationships between chemical reactions and electricity is known as electrochemistry. A chemical reaction in which current is either externally supplied or it is produced due to a spontaneous chemical reaction is known as an electrochemical reaction.
- The electrode potentials help us to determine the feasibility of a reaction. Generally, for measuring the potential of an electrode we use a Standard Hydrogen Electrode as it has a potential of zero.
We are given in the question that
$E_{A{u^ + }/Au}^0 = 1.69{\text{V}}$, it represents the reaction:
$A{u^ + } + {e^ - } \to Au(s)$
$E_{A{u^{3 + }}/Au}^0 = 1.40V$, which denotes the reaction:
$A{u^{3 + }} + 3{e^ - } \to Au(s)$
We need to find $E_{A{u^{3 + }}/A{u^ + }}^0$, for which the reaction is given as:
$A{u^{3 + }} + 2{e^ - } \to A{u^ + }$
So we can say that $2E_{A{u^{3 + }}/A{u^ + }}^0 = \left( {3 \times E_{A{u^{3 + }}/Au}^0} \right) - E_{A{u^ + }/Au}^0$
Substituting the values:
$ \Rightarrow E_{A{u^{3 + }}/A{u^ + }}^0 = \dfrac{{\left( {3 \times E_{A{u^{3 + }}/Au}^0} \right) - E_{A{u^ + }/Au}^0}}{2} = \dfrac{{3 \times 1.69 - 1.4}}{2}$
$\therefore E_{A{u^{3 + }}/A{u^ + }}^0 = 1.255V$

Thus, the correct answer is C.

Note: The standard electrode potentials of various metals and elements are listed in an electrochemical series. An electrochemical series is the list of reduction potentials of different electrodes arranged in an increasing order. The species that is present higher in the series is a stronger reducing agent and the species present lower in the series is a stronger oxidizing agent. A given redox reaction is possible only if the species that has a higher reduction potential is reduced.