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Mg +2 Ag + (0.0001M) → Mg 2+(0.130M) + 2Ag, E° =3.17V |
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Answer» ong>Answer: ANSWER Mg+2Ag+(0.0001M)⟶Mg 2+ (0.130M)+2Ag
=E o − 2 0.0591
LOG[ (AG + ) 2
(Mg 2+ )
] E cell
=8.17− 2 0.0591
log[ −8
0.130
]=2.959 V |
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