J.R. S. answered 11/17/23
Ph.D. University Professor with 10+ years Tutoring Experience
Oxidation takes place at the anode and reduction takes place at the cathode.
A. Half reaction at the anode:
Sn(s) ==> Sn2+(aq) + 2e- (oxidation at the anode)
B. Half reaction at the cathode:
Ag+(aq) + e- ==> Ag(s) (reduction at the cathode)
C. Balanced cell reaction (multiply reduction reaction by 2 to equalize electrons being transferred)
Sn(s) + 2Ag+(aq) ==> Sn2+(aq) + 2Ag(s)