Question

13. Pb2+ 2e-->Pb(s),E' =-0.13 V
Sn^2 + 2e→ Sn(s), E = -0.16 V
Ni2+ -->Ni(s),E= -0.25 V
Cr*3 + 3e -->Cr(s), E = -0.74 V
Based on the above data, the reducing power of Pb, Sn, Ni and Cr is in the order
(1) Pb > Sn > Ni > Cr
(3) Cr > Sn> Ni > Pb
(2) Cr > Ni > Sn > Pb
(4) Sn > Ni > Cr > Pb
​

Answer 1

Explanation:

The cell potential in Chapter 17.2 Galvanic Cells (+0.46 V) results from the ... anode\;(oxidation):Cu(s)Cu2+(aq)+2e−cathode\;(reduction):2Ag+(aq)+2e−2Ag( s)