The standard electrode potentials of two half-cells are given as: $Ni^{2+}_{(aq)} + 2e^{-} \rightarrow Ni_{(s)}$; $E^o = -0.25 \ V$ and $Zn^{2+}_{(aq)} + 2e^{-} \rightarrow Zn_{(s)}$; $E^o = -0.76 \ V$. What is the standard $emf$ of the cell formed by connecting these two half-cells in $volt$?

The $E^{0}_{Red}$ values for $P, Q, R$ and $S$ are $-2.90 \, V, +0.34 \, V, +1.20 \, V$ and $-0.76 \, V$ respectively. The decreasing order of their reactivity is:

Given the standard reduction potentials at $25\,^oC$ for the following half-reactions:
$Zn^{2+}_{(aq)} + 2e^- \rightleftharpoons Zn_{(s)}, E^o = -0.76\,V$
$Cr^{3+}_{(aq)} + 3e^- \rightleftharpoons Cr_{(s)}, E^o = -0.74\,V$
$2H^+_{(aq)} + 2e^- \rightleftharpoons H_{2(g)}, E^o = 0.00\,V$
$Fe^{3+}_{(aq)} + e^- \rightleftharpoons Fe^{2+}_{(aq)}, E^o = +0.77\,V$
Which of the following is the strongest reducing agent?

The reducing ability of the metals $K$,$Au$,$Zn$ and $Pb$ follows the order

For the cell reaction:
$2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightarrow 2 Fe^{2+}_{(aq)} + I_{2(aq)}$
$E^{\ominus}_{cell} = 0.24 \ V$ at $298 \ K$. The standard Gibbs energy $(\Delta_r G^{\ominus})$ of the cell reaction in $kJ \ mol^{-1}$ is:
[Faraday constant $F = 96500 \ C \ mol^{-1}$]

The standard hydrogen electrode has a zero electrode potential because .....