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Question: If \({E^ \circ }_{\dfrac{{A{u^{2 + }}}}{{Au}}}\)is \(1.69V\) and \({E^ \circ }_{\dfrac{{A{u^{3 + }}}...

If E∘Au2+Au{E^ \circ }_{\dfrac{{A{u^{2 + }}}}{{Au}}}is 1.69V1.69V and E∘Au3+Au{E^ \circ }_{\dfrac{{A{u^{3 + }}}}{{Au}}} is 1.40V1.40V, then E∘Au2+Au+{E^ \circ }_{\dfrac{{A{u^{2 + }}}}{{A{u^ + }}}}
A.0.19V0.19V
B.2.945V2.945V
C.1.255V1.255V
D.None

Explanation

Solution

In electrochemistry, electrode potential is the electromotive force of a galvanic cell built from a standard reference electrode and another electrode to be characterized. By convention, the reference electrode is the standard hydrogen electrode. It is defined to have a potential of zero volts.

Complete answer:
The Nernst equation is an equation that relates the reduction potential of a reaction to the standard electrode potential, temperature, and activities of the chemical species undergoing reduction and oxidation.
The overall cell potential can be calculated by using the equation E∘Cell=E∘Reduction−E∘Oxidation{E^ \circ }_{Cell} = {E^ \circ }_{Reduction} - {E^ \circ }_{Oxidation}. Before adding the two reactions together, the number of electrons lost in the oxidation must equal the number of electrons gained in the reduction. The silver half-cell reaction must be multiplied by two
⇒Ecell=Eθcell−RTzFln⁡Qr\Rightarrow {E_{cell}} = {E^\theta }_{cell} - \dfrac{{RT}}{{zF}}\ln {Q_r}
⇒Au++e−→Au\Rightarrow A{u^ + } + {e^ - } \to AuConsider this as equation (I)(I)
⇒Au→Au3++3e−\Rightarrow Au \to A{u^{3 + }} + 3{e^ - }Consider this as equation(II)(II)
⇒Au→Au3++3e−\Rightarrow Au \to A{u^{3 + }} + 3{e^ - } Consider this as equation(III)(III)
⇒ΔG∘(III)=ΔG∘(I)+ΔG∘(III)\Rightarrow \Delta {G^ \circ }_{(III)} = \Delta {G^ \circ }_{(I)} + \Delta {G^ \circ }_{(III)}
⇒2×E∘cell×F=3×1.40F−F×1.69\Rightarrow 2 \times {E^ \circ }_{cell} \times F = 3 \times 1.40F - F \times 1.69
⇒−2×E∘cell=3×1.40−1.69\Rightarrow - 2 \times {E^ \circ }_{cell} = 3 \times 1.40 - 1.69
⇒−2E∘cell=4.2−1.69\Rightarrow - 2{E^ \circ }_{cell} = 4.2 - 1.69
⇒−E∘cell=2.512\Rightarrow - {E^ \circ }_{cell} = \dfrac{{2.51}}{2}
⇒−E∘cell=1.255V\Rightarrow - {E^ \circ }_{cell} = 1.255V
⇒E∘Au2+Au+=1.255V\Rightarrow {E^ \circ }_{\dfrac{{A{u^{2 + }}}}{{A{u^ + }}}} = 1.255V
So, the correct answer is C)1.255VC)1.255V

Additional information:
The factors affecting the magnitude of electrode potential is
Nature of metal or electrode
The concentration of metal ions in solution
Temperature
The value of electrode potential developed on an electrode also depends on
The concentration of ions in solution
The temperature of the system
Chemical nature of the metal or nonmetal
A number of electrons transferred in the half cell reactions.
It does not depend on the electrode.

Note:
The reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction, E∘Oxidation=−E∘Re⁡duction{E^ \circ }_{Oxidation} = - {E^ \circ }_{\operatorname{Re} duction}. Add the potentials of the half-cells to get the overall standard cell potential.