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 $E^{\circ}_{\text{cell}}$ for $Cd_{(s)} | Cd_{(1M)}^{2+} || Ag_{(1M)}^{+} | Ag_{(s)}$ is $1.2 \ V$,what is the emf of the cell at $25^{\circ} C$ (in $V$)?

For the galvanic cell,
$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$
$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$
$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$
$[\frac{2.303 \ RT}{F} = 0.059 \ V]$

Consider the cell whose $emf$ is $1.01 \ V$.
$Pt, H_2(1 \ atm) | H^{+}(pH = 4) || Ag^{+}(xM) | Ag$
What is the value of $x$? (Given: $E^o_{Ag^{+}|Ag} = +0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

In a cell that utilises the reaction $Zn_{(s)} + 2H^{+}_{(aq)} \to Zn^{2+}_{(aq)} + H_{2_{(g)}}$,addition of $H_2SO_4$ to the cathode compartment will:

$1 \ F$ electricity was passed through $Cu^{2+} (1.5 \ M, 1 \ L) / Cu$ and $0.1 \ F$ was passed through $Ag^{+} (0.2 \ M, 1 \ L) / Ag$ electrolytic cells. After this,the two cells were connected to make an electrochemical cell. The $emf$ of the cell thus formed at $298 \ K$ is:
Given: $E^0_{Cu^{2+} / Cu} = 0.34 \ V$,$E^0_{Ag^{+} / Ag} = 0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$ (in $V$)