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

For a cell having a standard $emf$ of $0.295 \ V$ at $25^o \ C$ involving a two-electron change,the equilibrium constant for the reaction is:

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. $A$ simple model for such a concentration cell involving a metal $M$ is:
$M_{(s)} \mid M^{+}(aq; 0.05 \ M) \parallel M^{+}(aq; 1 \ M) \mid M_{(s)}$
For the above electrolytic cell the magnitude of the cell potential $|E_{cell}|=70 \ mV$.
$1.$ For the above cell
$(A)$ $E_{cell} < 0 ; \Delta G > 0$ $(B)$ $E_{cell} > 0 ; \Delta G < 0$
$(C)$ $E_{cell} < 0 ; \Delta G^{\circ} > 0$ $(D)$ $E_{cell} > 0 ; \Delta G^{\circ} > 0$
$2.$ If the $0.05 \ M$ solution of $M^{+}$ is replaced by $0.0025 \ M$ $M^{+}$ solution,then the magnitude of the cell potential would be
$(A)$ $35 \ mV$ $(B)$ $70 \ mV$ $(C)$ $140 \ mV$ $(D)$ $700 \ mV$
Give the answer for questions $1$ and $2$.

What will be the $EMF$ for the given cell?
$Pt | H_2(g, P_1) | H^{+}(aq.) (1 \ M) || H^{+}(aq.) (1 \ M) | H_2(g, P_2) | Pt$

The cell potential for the following cell notation is approximately
$M_{(s)} | M^{3+}(aq, 0.01 \ M) || N^{2+}(aq, 0.1 \ M) | N_{(s)}$
$E_{M^{3+} / M}^0 = 0.6 \ V$ and $E_{N^{2+} / N}^0 = 0.1 \ V$ (in $V$)

For a cell involving a one-electron change at $25^o C$,$E^{o}_{cell} = 0.591 \ V$. The equilibrium constant for the reaction is .....