Find Ecell° for the reaction: 3Zn(.s.)+2(Cr)+ 3(.aq..) arrow 3(Zn)+ 2(.aq..)+2Cr(.s.) Given: E((Zn)+ 2 / Zn)o=-0.76V;E(Cr)+ 3 / Cr0=-0.74V

Question

Find Ecell° for the reaction: 3Zn(.s.)+2(Cr)+ 3(.aq..) arrow 3(Zn)+ 2(.aq..)+2Cr(.s.) Given: E((Zn)+ 2 / Zn)o=-0.76V;E(Cr)+ 3 / Cr0=-0.74V (A) 0.02V (B) 0.08V (C) -1.5V (D) 1.5V

Tardigrade Admin · Accepted Answer

Ecell0=ECr3 + / Cr0-EZn2 + / Zn0

Q. Find $E_{ce ll}^{\circ}$ for the reaction :
$3 Z n (s) + 2 (C r)^{+ 3} (a q .) \to 3 (Z n)^{+ 2} (a q .) + 2 C r (s)$
Given : $E_{((Z n)^{+ 2} / Z n)}^{o} = - 0.76 V; E_{(C r)^{+ 3} / C r}^{0} = - 0.74 V$

$0.02 V$

$0.08 V$

$- 1.5 V$

$1.5 V$

$E_{ce ll}^{0} = E_{C r^{3 +} / C r}^{0} - E_{Z n^{2 +} / Z n}^{0}$

Q. Find Ecell∘​ for the reaction : 3Zn(s)+2(Cr)+3(aq.)→3(Zn)+2(aq.)+2Cr(s) Given : E((Zn)+2/Zn)o​=−0.76V;E(Cr)+3/Cr0​=−0.74V

0.02V

0.08V

−1.5V

1.5V