The electrode potentials for Cu2+(aq)+e--Cu+(aq) and Cu+(aq)+e--Cu(s) are + 0.15 V and + 0.50 V, respectively. The value of E°Cu2+/Cu will be :

Question

The electrode potentials for Cu2+(aq)+e-->Cu+(aq) and Cu+(aq)+e-->Cu(s) are + 0.15 V and + 0.50 V, respectively. The value of E°Cu2+/Cu will be : (A) 0.500 V (B) 0352 V (C) 0.650 V (D) 0.150 V

Tardigrade Admin · Accepted Answer

cu2++le- arrow Cu+ ; Δ G°1=-n1E1°F] (cu++le- arrow Cu ; Δ G° 2=-n2E2° F/cu2++2e- arrow Cu ; Δ G°=Δ G° 1=Δ G°2) -nEÂ°F=1 n1 EÂ°F+(-1)n2 EÂ°2F -nEÂ° F=-F(n1 EÂ°1+n2 EÂ°2F) EV=(n1EÂ°1+n2EÂ°2/n)=(0.15×1+0.50×1/2)=0.325

Q. The electrode potentials for
$Cu X^{2 +} (aq) X^{+} e X^{-} Cu X^{+} (aq)$ and $Cu X^{+} (aq) X^{+} e X^{-} Cu (s)$
are $+ 0.15 V$ and $+ 0.50 V$ , respectively. The value of $E_{C u^{2 +} / C u}^{°}$ will be :

$0.500 V$

$0352 V$

$0.650 V$

$0.150 V$

Q. The electrode potentials for CuX2+(aq)X+eX−​CuX+(aq) and CuX+(aq)X+eX−​Cu(s) are +0.15V and +0.50V, respectively. The value of ECu2+/Cu°​ will be :

0.500V

0352V

0.650V

0.150V

cu2++le−→Cu+;ΔG1∘​=−n1​E1∘​F] cu2++2e−→Cu;ΔG∘=ΔG1∘​=ΔG2∘​cu++le−→Cu;ΔG2∘​=−n2​E2∘​F​ −nE°F=1n1​E°F+(−1)n2​E2°​F −nE°F=−F(n1​E1°​+n2​E2°​F) EV=nn1​E1°​+n2​E2°​​=20.15×1+0.50×1​=0.325

Q. The electrode potentials for
$Cu X^{2 +} (aq) X^{+} e X^{-} Cu X^{+} (aq)$ and $Cu X^{+} (aq) X^{+} e X^{-} Cu (s)$
are $+ 0.15 V$ and $+ 0.50 V$ , respectively. The value of $E_{C u^{2 +} / C u}^{°}$ will be :

$0.500 V$

$0352 V$

$0.650 V$

$0.150 V$