Question

\[\begin{array}{l}
C{u^ + } + {e^ - } \to Cu\,;\,{E^ \circ } = {x_1}\,volt\\
C{u^{ + 2}} + 2{e^ - } \to Cu\,;\,{E^ \circ } = {x_2}\,volt,\,then\\
C{u^{ + 2}} + {e^ - } \to C{u^ + }\,;\,{E^ \circ }(volt)\,will\,be\\
\end{array}\]
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.