The voltage of the cell, $\mathrm{Pb}\left(\mathrm{s}\right) \left| {\mathrm{PbSO}}_4\left(\mathrm{s}\right) \right| {\mathrm{NaHSO}}_4 \left(0.6\mathrm{M}\right) \left|\right| {\mathrm{Pb}}^{2+} \left(2.5\times {10}^{-5}\right) | \mathrm{Pb}\left(\mathrm{s}\right)$ $\text{is} 0.061 \mathrm{V}. \text{Calculate }{\mathrm{K}}_2=\frac{\left[{\mathrm{H}}^{+}\right] \left[{\mathrm{SO}}_4^{2-}\right]}{\left[{\mathrm{HSO}}_4^{-}\right]}.$ If ${\mathrm{K}}_2 =\mathrm{a}\times {10}^{-2}$, report $\text{'}\mathrm{a}\text{'}$ correct up to one place of decimal.

$\text{Given,} {\mathrm{PbSO}}_4\left(\mathrm{s}\right)+2{\mathrm{e}}^{-}\to \mathrm{Pb}\left(\mathrm{s}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right); {\mathrm{E}}^0=-0.356 \mathrm{V}$ and ${\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Pb}\left(\mathrm{s}\right); {\mathrm{E}}^0=-0.126 \mathrm{V}$.

Determine the emf of the following cell:

$\mathrm{Pb} \left| {\mathrm{PbSO}}_4\left(\mathrm{s}\right), {\mathrm{SO}}_4^{2-}\left(1.0 \mathrm{M}\right) \Vert {\mathrm{H}}^{+}\left(1.0 \mathrm{M}\right) \right| {\mathrm{H}}_2\left(1.0 \mathrm{atm}\right) | \mathrm{Pt}$.

The anode is essentially a lead electrode, $\mathrm{Pb}|{\mathrm{Pb}}^{2+} \left(\mathrm{aq}\right)$. However, the anode solution is saturated with ${\mathrm{PbSO}}_4$ so that, ${\mathrm{Pb}}^{2+}$ ion concentration is determined by the solubility product of ${\mathrm{PbSO}}_{4 }\left(=1.7\times {10}^{-8}\right).$ $\left({\mathrm{E}}_{{\mathrm{Pb}}^{2+}|\mathrm{Pb}}^{\circ }= –0.13 \mathrm{V}, \log 1.7=0.23\right)$

Round off your answer up to two places of decimals.

Find the solubility product of a saturated solution of ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$ in water at $298 \mathrm{K}$ if the emf of the cell $\mathrm{Ag}\left|{\mathrm{Ag}}^{+}\left(\text{satd. }{\mathrm{Ag}}_2{\mathrm{CrO}}_4\text{ soln. }\right)\right|\left|{\mathrm{Ag}}^{+}\right(0.1 \mathrm{M}\left)\right|\mathrm{Ag}$ is $0.164 \mathrm{V}$ at $298 \mathrm{K}$. If solubility product is $\mathrm{a}\times {10}^{-12}$, report $\text{'}\mathrm{a}\text{'}.$

Calculate emf of $\mathrm{Ag}|\mathrm{AgCl}$ electrode immersed in $1 \mathrm{M}$ $\mathrm{KCl}$ at $25°\mathrm{C}$. ${\mathrm{K}}_{\mathrm{sp}}\left(\mathrm{AgCl}\right) = 1.8\times {10}^{-10}, \mathrm{E}{°}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}} = 0.799 \mathrm{V}$.

(mention the answer upto three decimal places after rounding off)

The emf of the cell $\mathrm{Ag}\left| \mathrm{AgCl}, \mathrm{KCl} (0.05 \mathrm{M}) \Vert {\mathrm{AgNO}}_3 (0.05 \mathrm{M})\right| \mathrm{Ag}$ is $0.788 \mathrm{V}$. Find ${\mathrm{K}}_{\mathrm{sp}}$ for $\mathrm{AgCl}$.

If the answer is of type $\mathrm{y}\times {10}^{-16}$, report the value of $\mathrm{y}$ correct up to one place of decimal.

An excess of liquid mercury was added to a ${10}^{-3} \mathrm{M}$ acidic solution of ${\mathrm{Fe}}^{3+}$. It was found that only $4.6\%$ of the iron remained as ${\mathrm{Fe}}^{3+}$ at equilibrium at $25°\mathrm{C}$. Calculate ${{\mathrm{E}}^0}_{{\mathrm{Hg}}_2^{2+},\mathrm{Hg}}.$ Assume that the only reaction taking place is

Given that $2\mathrm{Hg}+2{\mathrm{Fe}}^{3+}={\mathrm{Hg}}_2^{2+}+2{\mathrm{Fe}}^{2+}$

${\mathrm{E}}^{\mathrm{o}}$_Fe3+,Fe2+= 0.771volt

$\mathrm{Ag}\left(\mathrm{s}\right)\left|{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\right|\mathrm{KCl}\left(\mathrm{aq}\right)\left|\mathrm{AgCl}\left(\mathrm{s}\right)\right|\mathrm{Ag}\left(\mathrm{s}\right)$

Find the value of $\left|{\mathrm{E}}_{\mathrm{cell} }°\right|$ in volt. [$1 \mathrm{F}=96500$ coulomb]

[Given: ${\mathrm{\Delta }}_{\mathrm{f}}\mathrm{G}°\left(\mathrm{AgCl},\mathrm{s}\right)=-109\mathrm{kJ}/\mathrm{mol}, {\mathrm{\Delta }}_{\mathrm{f}}\mathrm{G}°\left({\mathrm{Cl}}^{-},\mathrm{aq}\right)=-129\mathrm{kJ}/\mathrm{mol}$, ${\mathrm{\Delta }}_{\mathrm{f}}\mathrm{G}°\left({\mathrm{Ag}}^{+},\mathrm{aq}\right)=77\mathrm{kJ}/\mathrm{mol}$]

One beaker contains a $100\mathrm{ml}$ solution of $0.04\mathrm{M} {\mathrm{KMnO}}_4, 0.01\mathrm{M} {\mathrm{MnSO}}_4$, and $1.00\mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$; and a second beaker containing $100\mathrm{ml}$ solution of $0.30\mathrm{M} {\mathrm{FeSO}}_4$ and $0.03\mathrm{M} {\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3$ are mixed in a large container. After the allowed time, beaker is connected to SHE by a salt bridge, and platinum electrodes are placed in each.The electrodes are connected via a wire with a voltmeter in between. What would be the single electrode potential of mixture relative to SHE?

${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}^0=0.771 \mathrm{V}, {\mathrm{E}}_{{\mathrm{MnO}}_4^{-}/{\mathrm{Mn}}^{2+}}^0=1.51 \mathrm{V}$

$\log \frac{0.0625}{0.1316}=-0.32$ and $\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059$