1.     a) Na
        b) Na
        c) Fe(II)
2.     a) Ni(II) + Cd --> Ni + Cd(II)  (E > 0)
        b) E = -0.25 V + 0.403 V = 0.153 V
        c) oxidation: Cd
        d) reduction: Ni(II)
        e) oxidation at the anode, reduction at the cathode; electron through the wire from Cd to Ni(II);  Bromide toward Cd
        f) E = E^o - (RT/nF)lnQ
               = 0.153 V - (0.0592/2)ln 0.01/1
               = 0.29 V