Question

For a given electrochemical cell:


Given that:
$A\left(s\right)\rightleftharpoons A^{n+}\left(aq\right)+n{e}^{-}$
$E^{\circ }=-1.5V$

$B\left(s\right)\rightleftharpoons B^{n+}\left(aq\right)+n{e}^{-}$
$E^{\circ }=-3.1V$

Choose the correct statement.

• A. Both A and B will get reduced
• B. Both A and B will get oxidised
• C. A gets oxidised and B gets reduced.
• D. A gets reduced and B gets oxidised

Answer 1

The correct option is C A gets oxidised and B gets reduced.

For a working electrochemical cell, one of the components will get reduced and the other one will get oxidised.
Reaction 1 :
$A\left(s\right)\rightleftharpoons A^{n+}\left(aq\right)+n{e}^{-}$
$E^{\circ }=-1.5V$

Reaction 2:
$B\left(s\right)\rightleftharpoons B^{n+}\left(aq\right)+n{e}^{-}$
$E^{\circ }=-3.1V$
Since for reaction 2, $E^{\circ }$ has a lower value than that of reaction 1, so forward reaction for reactant B will be less spontaneous than the forward reaction of reactant A.
Consequently reactant B has a tendency to reduce more as compared to reactant A.
Hence, A gets oxidised and B gets reduced.