Calculate the cell potential for the cell $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.85 \ M) | Cd_{(s)}$ at $298 \ K$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $V$)

The correct representation of Nernst's equation for the reduction of a metal ion $M^{n+}$ to metal $M$ is:

For the cell,$Mn_{(s)} | Mn^{2+}_{(aq)} (0.4 \, M) || Sn^{2+}_{(aq)} (0.04 \, M) | Sn_{(s)}$,calculate the free energy change $(\Delta G)$ at $298 \, K$ in $kJ$. Given: $E^o_{Mn^{2+}/Mn} = -1.18 \, V$; $E^o_{Sn^{2+}/Sn} = -0.14 \, V$; $\frac{2.303RT}{F} = 0.06$.

The Nernst equation is related to:

Calculate the cell potential for $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.2 \ M) | Cd_{(s)}$ at $298 \ K$. (in $V$)

The standard Gibbs energy change in $kJ \ mol^{-1}$ for a galvanic cell $A_{(s)} + B_{(aq)}^{3+} \longrightarrow A_{(aq)}^{3+} + B_{(s)}$ that has a standard emf of $0.5 \ V$ is: $\left(F = 96500 \ C \ mol^{-1}\right)$