I can't read your equation. The idea here is that Al goes into solution and Ag^+ comes out of solution. You can write the two half cells.
Al ==> Al^+3 + 3e Eo = ??
Ag^+ +e ==> Ag(s) Eo = ??
Then balance the equations and add the two Eo values.
Tarnished silver contains Ag2S. The tarnish can be removed by placing silverware in an aluminum pan containing an inert electrolyte soln, such as NaCL. Explain the electrochemical principle for this procedure, [The standard reduction potential for the half-cell reaction Ag2S(s)+2e-¡æ2Ag(s)+s^2-(aq) is .0.72V]
1 answer
1 answer