In acidic medium,$MnO_4^-$ acts as an oxidising agent: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$. If the $H^+$ ion concentration is doubled,the electrode potential of the half-cell will:

What is the reduction potential of a silver wire dipped in a $0.1 \ M \ AgNO_3$ solution at $25^\circ C$?

The cell reaction involving the quinhydrone electrode is given by the following equation:
$C_6H_4(OH)_2 \rightleftharpoons C_6H_4O_2 + 2H^+ + 2e^-$,$E^{\circ} = 1.30 \ V$
What will be the electrode potential at $pH = 3$ (in $V$)?

For a $Daniel$ cell $Zn | ZnSO_{4(0.01 \ M)} || CuSO_{4(1 \ M)} | Cu$ at $298 \ K$,the cell potential is $E_1$. When the concentrations of $ZnSO_4$ and $CuSO_4$ are changed to $1 \ M$ and $0.01 \ M$ respectively,the cell potential becomes $E_2$. Determine the relationship between $E_1$ and $E_2$.

Given $E^o _{Cr^{3+} / Cr} = -0.72 \, V$ and $E^o _{Fe^{2+} / Fe} = -0.42 \, V$. The potential for the cell $Cr \, | \, Cr^{3+}_{(0.1 \, M)} \, || \, Fe^{2+}_{(0.01 \, M)} \, | \, Fe$ is ......... $V$.

$Zn_{(s)} + Cl_2(1 \ atm) \to Zn^{2+} + 2Cl^-$. The $E^0_{cell}$ of the cell is $2.12 \ V$. To increase $E_{cell}$: