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Question: For the disproportionation of copper: \(2C{u^ + } \to C{u^{ + 2}} + Cu,{E^{\circ}}\) is: Given: ...

For the disproportionation of copper:
2Cu+→Cu+2+Cu,E∘2C{u^ + } \to C{u^{ + 2}} + Cu,{E^{\circ}} is:
Given:
E for Cu+2/CuC{u^{ + 2}}/Cu is 0.34V and
E for Cu+2/Cu+C{u^{ + 2}}/C{u^ + } is 0.15V
A. 0.49V
B. 0.19V
C. 0.38V
D. -0.38V

Explanation

Solution

This is a galvanic cell reaction. And in galvanic cell chemical energy is converted into electrical energy. The above given reaction in which copper electrode is reduced and oxidized so this above reaction is called disproportionation reaction.so we will find the electrode potential of each anode and cathode.

Complete answer:
We have Cu+2/CuC{u^{ + 2}}/C{u^{}} is 0.34 V0.34 \,V
This means Cu+2+2e−→CuC{u^{ + 2}} + 2{e^ - } \to Cu is a reduction reaction.
And we also have Cu+2/Cu+C{u^{ + 2}}/C{u^ + }
This means Cu+2+e−→Cu+C{u^{ + 2}} + {e^ - } \to C{u^ + } is also reduction reaction
∴\therefore The free energy change of a cell reaction is related to the electrical work that can obtained from the cell
This means ΔG=−nFEcell\Delta G = - nF{E_{cell}} and ΔG∘=−nFE∘\Delta {G^{\circ}} = - nF{E^{\circ}}
So,
\Delta {G^{\circ}}_{{\text{ }}C{u^{ + 2}} \to Cu} = \Delta {G^{\circ}}_{{\text{ C}}{{\text{u}}^{ + 2}} \to C{u^ + }} + \Delta {G^{\circ}}_{{\text{ }}C{u^ + } \to Cu}\\\
⇒−nFE∘Cu+2/Cu=−nFE∘Cu+2/Cu+−nFE∘Cu+/Cu\Rightarrow - nF{E^{\circ}}_{C{u^{ + 2}}/Cu} = - nF{E^{\circ}}_{C{u^{ + 2}}/C{u^ + }} - nF{E^{\circ}}_{C{u^ + }/Cu}
Putting all the values on the above equation
⇒2×0.34=1×0.15+E∘Cu+/Cu ⇒E∘Cu+/Cu=0.68−0.15 ⇒E∘Cu+/Cu=0.53 \Rightarrow 2 \times 0.34 = 1 \times 0.15 + {E^{\circ}}_{C{u^ + }/Cu} \\\ \Rightarrow {E^{\circ}}_{C{u^ + }/Cu} = 0.68 - 0.15 \\\ \Rightarrow {E^{\circ}}_{C{u^ + }/Cu} = 0.53
Now we have reactions:
Cu+→Cu+2+e− Cu++e−→Cu  C{u^ + } \to C{u^{ + 2}} + {e^ - } \\\ C{u^ + } + {e^ - } \to Cu \\\
E∘Cu+/Cu+2=−0.15V E∘Cu+/Cu=0.53V {E^{\circ}}_{C{u^ + }/C{u^{ + 2}}} = - 0.15V \\\ {E^{\circ}}_{C{u^{ + /Cu}}} = 0.53V
After combining both the reactions we get
2Cu+→Cu+2+Cu2C{u^ + } \to C{u^{ + 2}} + Cu
So overall electric potential for the reaction will be: 0.53−0.150.53 - 0.15
= 0.38 V0.38\, V

**Hence, The correct answer will be option number C

Note:**
1. Disproportionation reaction: A reaction in which oxidation and reaction takes place at the same time.
2. Oxidation reaction: Loss of electron in a reaction is called oxidation or increasing the oxidation number of an element is called the oxidation of the element.
3. Reduction reaction: Gain of the electron in a reaction is called reduction reaction or decreasing the oxidation number of an element in the reaction is called reduction reaction.