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

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2.96V

Step-by-step 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} - (2.303RT)/(nF) log \frac{ [M]}{ [{M}^(n + )] } \\ \\ \sf E_(cell) = E_(cell){ \degree} - (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 - (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 * {10}^(5) ] \\ \\ \sf \implies E = 3.17 - 0.0292 * 7.11 \\ \\ \sf \implies E = 3.17 - 0.2076 \\ \\ \sf \boxed {\pink \implies \pink{E = 2.96V}}`