Question

A voltaic cell consists of two half-cells. One half-cell contains a chromium electrode immersed in 1.00 M Cr(NO3)3 solution. The second half-cell contains a nickel electrode immersed in 1.00 M NI(NO3)2 solution. Nickel plates out on the nickel electrode as the voltaic cell runs. The beginning voltage of the cell is +0.487 V at 25°C. The standard electrode potential (standard reduction potential) of chromium at 25°C is -0.744 V. (a) Write a balanced half-reaction equation for the reaction occurring at the anode and a second balanced equation for the reaction occurring at the cathode. (Include states-of-matter under the given conditions in your answer.) anode: chemPad Help Greek cathode: chemPad Help Greek (b) Write a balanced equation to show the net reaction for the cell. (Include states-of-matter under the given conditions in your answer.) chemPad Help x.x = Greek

Answer 1

Ans - The cell consists of two half cells. Given that nickel plates out on the nickel N; 2+ is electrode, this means that seduced and this half reaction occurs at cathodo. To, Gr is oxidised at anode. @ Aus - Anodic half reaction - Cr(s) Gost (aq) + 3é Cathodic half reaction - Ni 2 Tag) + 2 é Ni (s) Now making the clutrons equal by multiplying :- st anodie reaction by 2 and cathodie by 3, Anodic half reaction & 26r(s) 23+ (aq) + be Cathodic half reaction - 3N;2+ (aq) + 6e 3 Ni (s). Ans - Adding the two half reactions, we get 26 (3) + 3 Ni 2+(aq) This is a balanced 2G3+ (aq) + 3 Ni (s). reaction for the cell. not

@ Ans - Given that fall = +0.987 v. Also, Estle = -0.749 v. EN; 2 / Ni =? We have, Ecell = Ecathode - Eanode - EN; 2/Ni - E G3tler -) 0.489 V = EN:*/No-4 0.744) v z) E N; ?/N; = (0.487 -0.744) = -0.257 v . Standard electrode potential for nickel is -0.257 V.