At what relative concentration of ${\mathrm{Zn}}^{2+}$ ions and ${\mathrm{Fe}}^{2+}$ ions will $\mathrm{Zn} \left(\mathrm{s}\right)$ and $\mathrm{Fe} \left(\mathrm{s}\right)$ have equal oxidation potential? If your answer is $\mathrm{y} \times {10}^{10}. \mathrm{then} \mathrm{y} \mathrm{is}$

 $$\begin{gathered} \left({\mathrm{E}}_{ \mathrm{Zn}, {\mathrm{Zn}}^{2+}}^0=0.76 \mathrm{V}, {\mathrm{E}}_{ \mathrm{Fe}, {\mathrm{Fe}}^{+2}}^0=0.44 \mathrm{V}\right) \\ \mathrm{antilog}(0.85) = 7.03 \end{gathered}$$

Calculate the total number of coulombs carried by ${\mathrm{Cl}}^{-}$ ions weighing $20.1 \mathrm{g}$.

If the answer is of type $\mathrm{y}\times {10}^4 \mathrm{C}$, report the value of $\mathrm{y}$ correct up to two places of decimals.

For the cells,

 $\begin{array}{l}2 \mathrm{Ag}+{\mathrm{Pt}}^{2+}\to 2 {\mathrm{Ag}}^{+}+\mathrm{Pt}; {\mathrm{E}}^0=0.4 \mathrm{V}\\ 2 \mathrm{Ag}+{\mathrm{F}}_2\to 2 {\mathrm{Ag}}^{+}+2{\mathrm{F}}^{-}; {\mathrm{E}}^0=2.07 \mathrm{V}\end{array}$

The potential for the reaction $\mathrm{Pt}\to {\mathrm{Pt}}^{2+}+2\mathrm{e}$ is assigned as zero. Determine the potential for the following electrodes.

${\mathrm{F}}^{-}\to \frac{1}{2}{\mathrm{F}}_2+\mathrm{e}$

For the cells 

$\begin{array}{l}2 \mathrm{Ag}+{\mathrm{Pt}}^{2+}\to 2 {\mathrm{Ag}}^{+}+\mathrm{Pt}; {\mathrm{E}}^0=0.4 \mathrm{V}\\ 2 \mathrm{Ag}+{\mathrm{F}}_2\to 2 {\mathrm{Ag}}^{+}+2 {\mathrm{F}}^{-}; {\mathrm{E}}^0=2.07 \mathrm{V}\end{array}$, 

The potential for the reaction, $\mathrm{Pt}\to {\mathrm{Pt}}^{2+}+2\mathrm{e}$ is assigned as zero. Determine the potential for the following electrodes in V.

$\mathrm{Ag}\to {\mathrm{Ag}}^{+}+\mathrm{e}$

Copper can reduce zinc ions if the resultant copper ions are at a sufficiently low concentration, by the formation of an insoluble salt. What is the maximum concentration of ${\mathrm{Cu}}^{2+}$ in the solution if this reaction is to occur when ${\mathrm{Zn}}^{2+}$ is $1 \mathrm{M}$?

The concentration of ${\mathrm{Cu}}^{+2}$ is $\mathrm{y}\times {10}^{-38}\mathrm{M}$. Find $\mathrm{y}$ correct up to one place of decimal.

Calculate the EMF of the cell at $25°\mathrm{C}$ in which the following reaction takes place. ${\mathrm{E}}_{{\mathrm{Zn}}^{2+}|\mathrm{Zn}}^{\circ }=-0.76 \mathrm{V}$ and ${\mathrm{E}}_{{\mathrm{Fe}}^{2+}|\mathrm{Fe}}^{\circ }=-0.44 \mathrm{V}$

$\mathrm{Zn}+{\mathrm{Fe}}^{2+} \left({10}^{-3} \mathrm{M}\right)\rightleftharpoons {\mathrm{Zn}}^{2+} \left({10}^{-4} \mathrm{M}\right)+\mathrm{Fe}$

Report your answer in Volts.