Question

C{u^ + } + {e^ - } \to Cu\,;\,{E^ \circ } = {x_1}\,volt\\\ C{u^{ + 2}} + 2{e^ - } \to Cu\,;\,{E^ \circ } = {x_2}\,volt,\,then\\\ C{u^{ + 2}} + {e^ - } \to C{u^ + }\,;\,{E^ \circ }(volt)\,will\,be\\\ \end{array}$$ A. $${x_1} - 2{x_2}$$ B. $${x_1} + 2{x_2}$$ C. $${x_1} - {x_2}$$ D. $$2{x_2} - {x_1}$$

Answer 1

Hint: The electrode potential of a reaction is independent of any linear factors. If we multiply an equation but any linear factor, its electrode potential remains unchanged.

Complete step by step answer:
Electrode potential, E, in chemistry or electrochemistry, is the electromotive force of a cell built of two electrodes. On the left-hand side of the cell diagram is the standard hydrogen electrode (SHE) and on the right-hand side is the electrode in question. The standard hydrogen electrode has a potential of 0 V. Thus, the electrode potential of the cell is equal to the electrode potential of the right electrode. The SHE (Standard Hydrogen Electrode) is the anode (left) and an electrode is a cathode (right).
In the given question, three reactions and their electrode potentials are given. Let the change in Gibbs energy of the three reactions be $\Delta {G_1}$, $\Delta {G_2}$ and $\Delta {G_3}$ respectively.
$Cu^{+2} + e^- \longrightarrow Cu^+$; $\Delta {G_3}$
$Cu^{+} + e^- \longrightarrow Cu$; $\Delta {G_1}$

$Cu^{+2} + 2e^- \longrightarrow Cu$; $\Delta {G_2}$

Thus, reaction 2 is a sum of reaction 1 and reaction 3.
Let the electrode potential of the third reaction be $x_3$.
Therefore,
$\Delta {G_2}$ = $\Delta {G_1}$ + $\Delta {G_3}$
$\Rightarrow - nF{x_2} = - nF{x_1} + ( - nF{x_3})$
$\Rightarrow 2{x_2} = {x_1} + {x_3}$
$\Rightarrow {x_3} = 2{x_2} - {x_1}$

Hence, the correct answer is (D).

Note: Remember that we cannot directly add the reactions unless it is a cell reaction which involves no net electrons. For adding and comparing other reactions, we need to use the concept of Gibbs free energy and the electrons involved are accounted for.