(a) Write the reactions for the discharge and charge of a nickel–cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell:Cd(OH)2(s) + 2 e– Cd(s) + 2 OH–(aq) E°red = –0.76 VNiO(OH)(s) + H2O(l) + e Ni(OH)2(s) + OH–(aq) E°red = +0.49 V(c) A typical nicad voltaic cell generates an emf of +1.30 V. Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constantfor the overall nicad reaction based on this typical emf value.

Chemistry

answer

watching

213

views

periwinklealpaca516Lv1

4 Apr 2020

(a) Write the reactions for the discharge and charge of a nickel–cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell:
Cd(OH)₂(s) + 2 e^– Cd(s) + 2 OH^–(aq) E^°_red = –0.76 V
NiO(OH)(s) + H₂O(l) + e Ni(OH)₂(s) + OH^–(aq) E°_red = +0.49 V
(c) A typical nicad voltaic cell generates an emf of +1.30 V. Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant
for the overall nicad reaction based on this typical emf value.

answer

watching

213

views

For unlimited access to Homework Help, a Homework+ subscription is required.

You have 0 free answers left.
Get unlimited access to 3.8 million step-by-step answers.

Get unlimited access

Already have an account? Log in

Beverley SmithLv2

26 May 2020

Unlock all answers

Get 1 free homework help answer.

Already have an account? Log in

Join us on Discord

Chemistry Study Group

Join now

Related questions

1) Calculate the value of the solubility product constant for PbSO₄ from the half-cell potentials.
PbSO₄(s) + 2e^- â Pb(s) + SO₄^2-(aq) EÂº = -0.397 V
Pb²⁺ + 2e^- â Pb(s) EÂº = -0.148 V

2) Use standard reduction potentials (shown below) to calculate the potential of a nickel-cadmium cell that uses a basic electrolyte that has a 0.6 M hydroxide concentration.
Cd(OH)₂(s) + 2e^- â Cd(s) + 2OH^-(aq) EÂº = -0.83 V
NiO(OH)s) + H₂O(l) + e^- â Ni(OH)₂(s) + OH^-(aq) EÂº = 0.52 V

In some applications nickel–cadmium batteries have been replaced by nickel–zinc batteries. The overall cell reaction for this relatively new battery is

H₂O (l) + 2 NiO(OH) (s) + Zn(s) →2 Ni(OH)₂(s) + Zn(OH)₂(s)

a)What is the cathode half-reaction? (b) What is the anode half-reaction? (c) A single nickel–cadmium cell has a voltage of 1.30 V. Based on the difference in the standard reduction potentials of Cd²⁺ and Zn²⁺, what voltage would you estimate a nickel–zinc battery will produce? (d) Would you expect the specific energy density of a nickel–zinc battery to be higher or lower than that of a nickel–cadmium battery?

peachtermite235

Heart pacemakers are often powered by lithium–silver chromate “button” batteries.

The overall cell reaction is

2 Li(s) + Ag₂CrO₄(s)→ Li₂CrO₄(s) + 2Ag(s)

(a) Lithium metal is the reactant at one of the electrodes of the battery. Is it the anode or the cathode? (b) Choose the two half-reactions from Appendix E that most closely approximate

the reactions that occur in the battery. What standard emf would be generated by a voltaic cell based on these half reactions? (c) The battery generates an emf of + 3.5 V. How close is this value to the one calculated in part (b)? (d) Calculate the emf that would be generated at body temperature, 37 °C. How does this compare to the emf you calculated in part (b)?

One Class Solution:-a) Li(s) is oxidized at the anode.

Ag₂CrO₄(s)+ 2e-→ 2Ag(s) + CrO₄^2-(aq), E= 0.446V
Li(s) → Li+(aq)+e^- E=–3.05V
2 Li(s) + Ag₂CrO₄(s)→ Li₂CrO₄(s) + 2Ag(s)

b) E⁰=0.446V-(-3.05V) =3.496=3.50V

c) The emf of the battery, 3.5 V, is exactly the standard cell potential calculated in part (b).

d) For this battery at ambient conditions, E⁰= E⁰, so logQ=O. This makes sense because all reactants and products in the battery are solids and thus present in their standard states. Assuming that E⁰ is relatively constant with temperature, the value of the second termin the Nernst equation is= 0 at 37°C, and E=3.5V.

cerisegoose810