Calculate the e.m.f. of the half-cell given below: $Fe | FeSO_4$ $(a = 0.1 \ M)$ where $E^o_{OP} = 0.44 \ V$. (in $V$)

If $E^{\circ}(Mg^{+2}_{(aq)} \mid Mg_{(s)}) = -2.37 \ V$. What is the potential for $Mg_{(s)} \rightarrow Mg^{+2}_{(0.01 \ M)} + 2 \overline{e}$ at $298 \ K$?

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: $\left(F = 96500 \ C \ mol^{-1}\right)$

For the redox reaction $Zn_{(s)} + Cu^{2+}(0.1 \ M) \to Zn^{2+}(1 \ M) + Cu_{(s)}$ taking place in a cell,$E_{cell}^o$ is $1.10 \ V$. $E_{cell}$ for the cell will be ............ $V$ $\left( 2.303 \frac{RT}{F} = 0.0591 \right)$

The $EMF$ of the following three galvanic cells are $E_1, E_2$ and $E_3$ respectively. Which of the following is correct?
$(i)$ $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (0.1 \ M) | Cu$
$(ii)$ $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (1 \ M) | Cu$
$(iii)$ $Zn | Zn^{2+} (0.1 \ M) || Cu^{2+} (1 \ M) | Cu$

For a cell reaction involving two electron changes,$E_{\text{cell}}^{\circ} = 0.3 \text{ V}$ at $25^{\circ}\text{C}$. The equilibrium constant of the reaction is: