Question

Which of the following metals, when coupled, will leads to a non-spontaneous cell reaction?
Given:
$A{u}^{+}\left(aq\right)+e^{-}\to Au\left(s\right);E^0=1.69V$
$C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right);E^0=0.34V$
$P{b}^{2+}\left(aq\right)+2e^{-}\to Pb\left(s\right);E^0=-0.13V$
$F{e}^{2+}\left(aq\right)+2e^{-}\to Fe\left(s\right);E^0=-0.44V$
$C{a}^{2+}\left(aq\right)+2e^{-}\to Ca\left(s\right);E^0=-2.87V$

• A. $Fe$ at anode and $Au$ at cathode
• B. $Pb$ at anode and $Au$ at cathode
• C. $Cu$ at anode and $Au$ at cathode
• D. $Cu$ at anode and $Ca$ at cathode

Answer 1

The correct option is D $Cu$ at anode and $Ca$ at cathode

Spontaneity and feasibility of the cell can be easily predicted from electrochemical series.
For spontaneous and feasible reactions of a cell, its cell potential must be positive.

To get a positive cell potential, the reduction potential of anode half cell should be lower than the reduction potential of cathode half cell.

In the given electrochemical,
$E^0$ value of $A{u}^{+}/Au$ is the highest. Hence, when a $A{u}^{+}/Au$ couple is used as cathode, the cell potential of the cell will be positive.

Thus, (a), (b) and (c) will leads to a spontaneous and feasible cell reaction.

Also,
$E^0$ value of $C{a}^{2+}/Ca$ couple is the least.
Hence, when it used as cathode, its cell potential value will be negative and the cell reaction will be non-spontaneous.