Questions in electrochemistry

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The emf of a Galvanic cell, with electrode potential of silver = + 0.80 V and that of copper = + 0.34 V, is
Given the half-cell reactions (i) $F{{e}^{2+}}(aq)+2{{e}^{-}}\to Fe\,(s);\,\,E=-0.44\,V$ (ii) $2{{H}^{+}}(aq)+\frac{1}{2}{{O}_{2}}(g)+2{{e}^{-}}\to {{H}_{2}}O\,(l);\,\,{{E}^{0}}=+1.23\,V$ ${{E}^{0}}$ for the reaction $Fe\,(s)+2{{H}^{+}}+\frac{1}{2}{{O}_{2}}(g)\to F{{e}^{2+}}(aq)+{{H}_{2}}O\,(l)$ is
Sodium is made by the elctrolysis of a molten mixture of about 40% NaCl and 60% $CaC{{l}_{2}}$ because
Standard electrode potentials of Zn and Fe are known to be (i) – 0.76 V and (ii) – 0.44 V respectively. How does it explain that galvanization prevents rusting of iron while zinc slowly dissolves away
Consider the given data. The numerical value of the standard cell potential for the reaction $2C{{r}^{3+}}+3C{{u}^{2+}}(aq) \rightleftharpoons 2C{{r}^{3+}}(aq)+3Cu\,(s)$ is
Calculate the equilibrium constant of the reaction, $C{{d}^{2+}}(aq)+Zn\,(s)\to Z{{n}^{2+}}(aq)+Cd\,(s)$ If $E_{C{{d}^{2+}}/Cd}^{0}=-0.403\,V$ and $E_{Z{{n}^{2+}}/Zn}^{0}=-0.763\,V$
Determine the electrode potential for
.$Zn+C{{u}^{2+}}(aq) \rightleftharpoons Cu+Z{{n}^{2+}}(aq)$, reaction quotient is $Q=\frac{[Z{{n}^{2+}}]}{[C{{u}^{2+}}]}$. Variation of ${{E}_{\text{cell}}}$ with log Q is of the type with OA = 1.10 V. ${{E}_{\text{cell}}}$ will be 1.1591 V when
For the half-cell: At pH = 2, electrode potential is
For the calomel half-cell. Hg, $H{{g}_{2}}C{{l}_{2}}|C{{l}^{-}}(aq)$ values of electrode potentials are plotted at different $\log [C{{l}^{-}}]$. Variation is represented by

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