Question

present the cell in which the following reaction place

Mg(s) + 2Ag+ (0.0001 M) ----> Mg2+ (0.130M) + 2Ag(s)

calculate its E if E∅ is 3.17 V​

Answer 1

$\Huge{\green}\fcolorbox{blue}{cyan}{\bf{\underline{\red{\color{red}Answer}}}}$

2.96V

Explanation:

$\sf Mg(s) + 2 {Ag}^{ + } (0.0001M) \longrightarrow {Mg}^{2 + } (0.130M) + 2Ag(s) \\ \\ \sf E{ \degree} = 3.17V$

As per the Nerest equation

$\sf E_{cell} = E_{cell}{ \degree} - \frac{2.303RT}{nF} log \frac{ [M]}{ [{M}^{n + }] } \\ \\ \sf E_{cell} = E_{cell}{ \degree} - \frac{0.0592}{n} log \frac{[M]}{[ {M}^{n + } ]}$

Here n = 2

at the n depicts the number of electrons

M = Mg as it is neutral after the product

while M+ = Ag2+ as it is positively charge after the product

$\implies \sf E =3.17 - \frac{0.0592}{2} log \frac{[Mg]}{ {[{Ag}^{ + }]}^{2} } \\ \\ \sf \implies E = 3.17 - 0.0292 log[ \frac{0.130}{ {10}^{ - 8} } ] \\ \\ \sf \implies E = 3.17 - 0.0292 log[130 \times {10}^{5} ] \\ \\ \sf \implies E = 3.17 - 0.0292 \times 7.11 \\ \\ \sf \implies E = 3.17 - 0.2076 \\ \\ \sf \boxed {\pink \implies \pink{E = 2.96V}}$