Question

The electrode potential, $E^{\circ }$ , for the reduction of $Mn{O}_4^{-}$ to $M{n}^{2+}$ in acidic medium is $+1.51V$. Which of the following metal(s) will be oxidised? The reduction reactions and standard electrode potentials for $Z{n}^{2+},A{g}^{+}$ and $A{u}^{+}$ are given as
$Z{n}_{\left(aq\right)}^{2+}+2e^{-}\to Z{n}_{\left(s\right)},E^{\circ }=-0.762V$
$A{g}_{\left(aq\right)}^{+}+e^{-}\rightleftharpoons A{g}_{\left(s\right)},E^{\circ }=+0.80V$
$A{u}_{\left(aq\right)}^{+}+e^{-}\rightleftharpoons A{u}_{\left(s\right)},E^{\circ }=+1.69V$

• A. Zn and Au
• B. Ag and Au
• C. Au
• D. Zn and Ag

Answer 1

Answer: (D)

More +ve the value of reduction potential more will be the tendency to undergo reduction.
The $E_{\text{red }}^0\left(Mn{O}_4^{-}\mid M{n}^{2+}\right)=+1.51V$ value is greater than $E_{\text{red }}^0\left(Z{n}^2+\mid Zn\right)=-0.762V$ and $E_{\text{red }}^0\left(A{g}^{+}\mid Ag\right)=+0.80V$. Thus, only $Zn$ and $Ag$ will be oxidised.