For the electrochemical cell, Mg ( s )left| M | vedclass

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Question

$\left.\left. Mg _{( s )} \mid Mg ^{+2} 	ext { (aq. } 1 M ight) \| Cu ^{+2} 	ext { (aq. } 1 M ight) \mid Cu _{( s )}$

$E = E ^0-\frac{ RT }{2

Vedclass · Accepted Answer

$10$

Vedclass · Answer

$\left.\left. Mg _{( s )} \mid Mg ^{+2} 	ext { (aq. } 1 M ight) \| Cu ^{+2} 	ext { (aq. } 1 M ight) \mid Cu _{( s )}$

$E = E ^0-\frac{ RT }{2 F } \ell n \left(\frac{\left[ Mg ^{+2}ight]}{\left[ Cu ^{+2}ight]}ight)$

$E = E ^0=2.70 	ext {, when }\left[ Mg ^{+2}ight]=\left[ Cu ^{+2}ight]$

$\left.\left. Mg _{( s )} \mid Mg ^{+2} 	ext { (aq, } x M ight) \| Cu ^{+2} 	ext { (aq. } 1 M ight) \mid Cu _{( s )}$

$Mg _{( s )}+ Cu _{( aq )}^{+2} ightarrow Cu _{( s )}+ Mg _{( aq )}^{+2}$

$E = E _0-\frac{ RT }{2 F } \ell n _{ e }\left(\frac{\left[ Mg ^{+2}ight]}{\left[ Cu ^{+2}ight]}ight)$

$2.67=2.70-\frac{300}{2 	imes 11500} \ln _e\left(\frac{ x }{1}ight)$

$-0.03=\frac{-300}{2 	imes 11500} \ell_e( x )$

$\ell n _{ e } x =\frac{-0.03 	imes 2 	imes 11500}{-300}=2.30$

$x=10.00$

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