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Question: For half reaction \(B\left( s \right) \to {B^{2 + }} + 2{e^ - }\) \(E_1^ \circ = - 0.44V\) And,...

For half reaction
B(s)→B2++2e−B\left( s \right) \to {B^{2 + }} + 2{e^ - } E1∘=−0.44VE_1^ \circ = - 0.44V
And, B2+→B3++e−{B^{2 + }} \to {B^{3 + }} + {e^ - } E2∘=1.3VE_2^ \circ = 1.3V
What is E∘{E^ \circ } for the reaction?
3e−+B3+→B(s)3{e^ - } + {B^{3 + }} \to B\left( s \right)
A.A. 0.86V0.86V
B.B. 0.14V0.14V
C.C. −0.14V - 0.14V
D.D. −0.28V - 0.28V

Explanation

Solution

A half cell reaction is either an oxidation reaction in which electrons are lost or a reduction reaction where electrons are gained. The reaction occurs in an electrochemical cell in which the electrons are lost at the anode through the oxidation and consumed at the cathode where the reduction occurs.

Complete step by step answer:
In the given question two half cell reactions are given. The value of We have to calculate the for the third half cell.
Given reaction,
B(s)→B2++2e−B\left( s \right) \to {B^{2 + }} + 2{e^ - } E1∘=−0.44VE_1^ \circ = - 0.44V
B2+→B3++e−{B^{2 + }} \to {B^{3 + }} + {e^ - } E2∘=1.3VE_2^ \circ = 1.3V
3e−+B3+→B(s)3{e^ - } + {B^{3 + }} \to B\left( s \right) E∘=?{E_ \circ } = ?
The value of E∘{E_ \circ } can be calculated using addition and subtraction of the half cell. Now,
E∘=−(E2∘+E1∘){E_ \circ } = - \left( {E_2^ \circ + E_1^ \circ } \right)
⇒\Rightarrow −nfE∘=−(−nfE2∘+(−nfE1∘)) - nf{E_ \circ } = - \left( { - nfE_2^ \circ + \left( { - nfE_1^ \circ } \right)} \right) [E∘=−nfE]\left[ {{E_ \circ } = - nfE} \right]
Where nn is the number of electrons and ff is the faraday constant.
⇒\Rightarrow −3×96500×E∘=−(−1×96500×1.3+2×96500×0.44) - 3 \times 96500 \times {E_ \circ } = - \left( { - 1 \times 96500 \times 1.3 + 2 \times 96500 \times 0.44} \right)
By solving above we get,
E∘=−0.14V{E_ \circ } = - 0.14V

Thus, the correct option is C.C.

Note:
The half–cell potential refers to the potential developed at the electrode of each half cell in an electrochemical cell. In an electrochemical cell, the overall potential is the total potentials of the half cells. The measurement of half-cell potential is used to evaluate:
Presence of corrosion, Potential vulnerability of element surface area to corrosion.
There are many types of half cell such as ion Gas ion-cell, Metal-Metal ion half cell, Metal amalgam ion cell half cell, Metal insoluble salt anion half cell and oxidation-reduction half cell.Oxidation takes place at anode and reduction at cathode in a cell of either type (Galvanic cell and electrolytic cell ). The main difference of the sign of the electrode. Table given below summarizes the sign convention for electrodes.