Solveeit Logo
Login

Question

Question: If \(1.\,C{u^{2 + }} + 2{e^ - } \to Cu,\,E^\circ = 0.337\,{\rm{V}}\\\\{\rm{2}}{\rm{.}}\,C{u^{2 + }} ...

If 1. Cu2++2e−→Cu, E∘=0.337 V2. Cu2++e−→Cu+, E∘=0.153 V1.\,C{u^{2 + }} + 2{e^ - } \to Cu,\,E^\circ = 0.337\,{\rm{V}}\\\\{\rm{2}}{\rm{.}}\,C{u^{2 + }} + {e^ - } \to C{u^ + },\,E^\circ = 0.153\,{\rm{V}}
Electrode potential E∘E^\circ for the reaction Cu++e−→CuC{u^ + } + {e^ - } \to Cu
A) 0.90 V
B) 0.30 V
C) 0.38 V
D) 0.52 V

Explanation

Solution

We know that electrode potential is the potential difference between the electrode and electrolyte. Electrode potential depends on the temperature and concentration of metal ion

Complete answer:
In a galvanic cell the reversible work done is equal to decrease in its Gibbs energy. Write an expression between the Gibbs free energy change and the standard cell potential.

ΔrG=−nFE∘{\Delta _r}G = - nFE^\circ ……(1)
Here, E represents the emf of the cell and nF is the amount of charge passed and ΔrG{\Delta _r}G is the Gibbs energy.
Given, electrode potential of Cu2++2e−→CuC{u^{2 + }} + 2{e^ - } \to Cu is 0.337 V0.337\,{\rm{V}} substitute the E∘E^\circ value in equation (1).
ΔrG1=−2×F×0.337 V{\Delta _r}{G_1} = - 2 \times F \times 0.337\,{\rm{V}}
E∘E^\circ value of Cu2++e−→Cu+C{u^{2 + }} + {e^ - } \to C{u^ + }is 0.153 V0.153\,{\rm{V}}substitute the E∘E^\circ value in equation (1).
ΔrG2=−1×F×0.153 V{\Delta _r}{G_2} = - 1 \times F \times 0.153\,{\rm{V}}
For Cu++e−→CuC{u^ + } + {e^ - } \to Cu
write the equation (1) as follows.
ΔrG3=−1×F×E∘{\Delta _r}{G_3} = - 1 \times F \times E^\circ
Write the relationship between ΔrG1,ΔrG2,ΔrG3{\Delta _r}{G_1} , {\Delta _r}{G_2} , {\Delta _r}{G_3} as follows:
ΔrG3=ΔrG1−ΔrG2{\Delta _r}{G_3} = {\Delta _r}{G_1} - {\Delta _r}{G_2}
Now, substitute the values in the above equation in terms of the standard cell potentials and calculate the electrode potential as follows:
ΔrG3=ΔrG1−ΔrG2 −1×F×E∘=(−2×F×0.337V)\-(−1×F×0.153 V) \-F×E∘=F(−0.674 V+0.153 V) E∘=0.521 V{\Delta _r}{G_3} = {\Delta _r}{G_1} - {\Delta _r}{G_2}\\\ - 1 \times F \times E^\circ = \left( { - 2 \times F \times 0.337{\rm{V}}} \right) \- \left( { - 1 \times F \times 0.153\,{\rm{V}}} \right)\\\ \- F \times E^\circ = F\left( { - {\rm{0}}{\rm{.674}}\,{\rm{V + }}0.153\,{\rm{V}}} \right)\\\ E^\circ = 0.521\,{\rm{V}}
Thus, the calculated E∘E^\circ value for reaction Cu++e−→CuC{u^ + } + {e^ - } \to Cuis 0.521 V0.521\,{\rm{V}}.
Hence, the correct answer is option ‘D’.

Additional information:
The concentrations of all the species involved in a half-cell is unity then the electrode potential is known as standard electrode potential. Redox reactions are based on electrochemical reactions and it comprises two half reaction reduction and oxidation. In an electrode, the standard electrode potential is determined by pairing it with the SHE and measuring the cell potential of the resulting galvanic cell.
Good oxidizing agents have high standard reduction potentials whereas good reducing agents have low standard reduction potentials. In redox reactions, key terms used to describe the reactants that pass electrons between reactants to form products are oxidising and reducing agents.

Note: Students get confused between the anode and the cathode electrodes. In electrochemistry, anode is positively charged electrode whereas cathode is negatively charged. Oxidation occurs at anode and reduction occurs at cathode.