Question

The standard electrode potential of $A{{g}^{+}}/Ag$ is $+0.80V$ and $C{{u}^{2+}}/Cu$ is $+0.34V$. These electrodes are connected through a salt bridge and if:
A. Copper electrode acts as a cathode, then ${{E}^{o}}_{cell}$ is $+0.46V$
B. Silver electrode acts as an anode, then ${{E}^{o}}_{cell}$ is $-0.34V$
C. Copper electrode acts as an anode, then ${{E}^{o}}_{cell}$ is $+0.46V$
D. Silver electrode acts as a cathode, then ${{E}^{o}}_{cell}$ is $-0.34V$

Answer 1

Hint: Recall what the standard electrode and reduction potentials signify. Which reaction will be carried out at the anode and which will be at the cathode according to this potential? Consider the Nernst equation to solve this question.

Complete step by step answer:
To find the total ${{E}^{o}}_{cell}$ we need to know the standard reduction potentials of the cathode as well as the anode. Let us first understand what reduction potentials are.
We know that when an atom of molecular species is reduced, it gains electrons. The reduction potential of any species defines its tendency to gain electrons. So, a species with a higher reduction potential will be more likely to gain electrons. We know that reduction occurs at the cathode, so when we identify the species that is going to get reduced, we will know which reaction will be carried out at the anode and which will be carried out at the cathode.
In both $A{{g}^{+}}/Ag$ and $C{{u}^{2+}}/Cu$, the species are going from a positive charge to becoming neutral. Thus, the values given to us are the standard reduction potentials. The standard reduction potential of $A{{g}^{+}}/Ag$ is $+0.80V$ and the standard reduction potential of $C{{u}^{2+}}/Cu$ is $+0.34V$. We can see that the reduction potential of silver is greater than that of copper. So, silver will get reduced and copper will get oxidized in this reaction. The reduction reaction takes place at the cathode, so the half cell potential of the cathode is $+0.80V$ and the half cell potential of the anode is $+0.34V$. The Nernst equation is defined as:

\[{{E}^{o}}_{cell}={{E}^{o}}_{cathode}-{{E}^{o}}_{anode}\]
We have deduced that, ${{E}^{o}}_{cathode}=+0.80V$ and ${{E}^{o}}_{anode}=+0.34V$, we will now substitute these values and obtain the answer.

\[\begin{align}
 & {{E}^{o}}_{cell}=(+0.80V)-(+0.34) \\
 & {{E}^{o}}_{cell}=+0.46V \\
\end{align}\]
So, according to what we have found out, the silver electrode will be the cathode, the copper electrode will be the anode and the cell potential will be $+0.46V$.
Hence, the correct answer is ‘C. Copper electrode acts as an anode, then ${{E}^{o}}_{cell}$ is $+0.46V$’
So, the correct answer is “Option C”.

Note: Always check whether the electrode potential values given are the oxidation potential values or the reduction potential values. Here, $A{{g}^{+}}/Ag$ so the silver with charge +1 is becoming neutral, so it is gaining electrons. This implies that it is getting reduced, and the reduction potential is given. If you want to convert the oxidation potential to the reduction potential, then ${{E}^{o}}(\text{standard reduction potential})=-{{E}^{o}}(\text{standard oxidation potential})$.