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Question

(A) The reduction half-reaction is:$MnO_4^- + 8H^{+} + 5e^- \rightarrow Mn^{2+} + 4H_2O$Using the Nernst equation for the electrode potential:$E = E^0

Vedclass · Accepted Answer

$3776$

Vedclass · Answer

(A) The reduction half-reaction is:$MnO_4^- + 8H^{+} + 5e^- ightarrow Mn^{2+} + 4H_2O$Using the Nernst equation for the electrode potential:$E = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-][H^{+}]^8}$Case $I$: When $[H^{+}] = 1 \, M$$E_1 = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-]}$Case $II$: When $[H^{+}] = 10^{-4} \, M$$E_2 = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-] 	imes (10^{-4})^8}$$E_2 = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-]} + \frac{0.059}{5} \log (10^{-4})^8$$E_2 = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-]} + \frac{0.059}{5} 	imes (-32)$Calculating the magnitude of change $|E_1 - E_2|$:$|E_1 - E_2| = \frac{0.059}{5} 	imes 32$$|E_1 - E_2| = 0.0118 	imes 32 = 0.3776 \, V$Given the change is $x 	imes 10^{-4} \, V$:$0.3776 \, V = 3776 	imes 10^{-4} \, V$Therefore,$x = 3776$.

Similar Questions

Calculate the $E_{cell}$ for $Zn_{(s)} | Zn^{2+}_{(0.1 \ M)} || Cr^{3+}_{(0.1 \ M)} | Cr_{(s)}$ at $25^{\circ} C$ if $E^{\circ}_{cell}$ is $0.02 \ V$. (in $V$)

For the cell reaction $Zn_{(s)} + 2H^+_{(aq)} \to Zn^{2+}_{(aq)} + H_{2(g)}$,what happens when $H_2SO_4$ is added to the cathode compartment?

Calculate the electrode potential at $298 \; K$ for $Zn|Zn^{2+}$ electrode in which the activity of zinc ions is $0.001 \; M$ and $E^o_{Zn^{2+}/Zn}$ is $-0.76 \; V$. (in $; V$)

At $298 \ K$, the $emf$ of the cell is ............ $V$.
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