4-A voltaic cell consists of an Mn/Mn²⁺half-cell and a Cd/Cd²⁺half-cell. Calculate {Cd²⁺}when {Mn²⁺}=4.37 M and E_cell= 0.704 V. Use reduction potential values of Mn²⁺= -1.18 V and for Cd²⁺= -0.40 V.

How manyamps are required to produce 94.4 g of lithium metal from a sampleof molten lithium chloride in 5.41 hours?

How many grams of thallium may be formed by the passage of 92,379 Cthrough an electrolytic cell that contains a molten Tl(I)salt.

answer

watching

263

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

Bunny GreenfelderLv2

17 Dec 2019

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

A voltaic cell consists of an Mn/Mn2+ half-cell and a Cd/Cd2+ half-cell. Calculate {Cd2+} when {Mn2+}= 0.11 M and Ecell = 0.775 V. Use reduction potential values of Mn2+ = -1.18 V and for Cd2+ = -0.40 V.

You are asked to construct a voltaic cell that would simulate an alkaline battery by providing an electrical output of 1.50 VV at the beginning of its discharge. After you complete it, your voltaic cell will be used to power an external device that draws a constant current of 0.50 AA (amperes) for 2.0 hh. You are given the following supplies: electrodes of each transition metal from manganese to zinc; the chloride salts of the +2 transition metal ions from Mn2+Mn2+ to Zn2+Zn2+ (MnCl2MnCl2, FeCl2FeCl2, CoCl2CoCl2, NiCl2NiCl2, CuCl2CuCl2, and ZnCl2ZnCl2); two 100 mLmL beakers; a salt bridge; a voltmeter; and wires to make electrical connections between the electrodes and the voltmeter. The standard reduction potentials for the divalent metal ions are as follows: Half-reactionMn2+(aq)Fe2+(aq)Co2+(aq)Ni2+(aq)Cu2+(aq)Zn2+(aq)++++++2e−2e−2e−2e−2e−2e−→→→→→→Mn(s)Fe(s)Co(s)Ni(s)Cu(s)Zn(s)E∘(V)−1.18−0.44−0.28−0.250.34−0.76 Since you were tasked with constructing a cell that produces a potential of 1.50 VV, you decide to fine tune your concentrations to minimize the error. If you used the same concentration for each electrolyte, your voltaic cell would produce a potential of 1.52 VV. Therefore, you should adjust the concentration ratio that would lower your cell potential by 0.02 VV. You and a fellow researcher decide that you should prepare two 100 mLmL solutions for the voltaic cell. Your fellow researcher already prepared one of the electrolyte solutions for you: 0.228 MM CuCl2 CuCl2. At what concentration should you prepare the MnCl2MnCl2 solution such that your initial voltaic cell potential is 1.50 VV? Assume that the temperature is consistently 25 ∘C∘C. If the battery is permitted to discharge for a period of two hours at a constant current of 0.50 AA, what will be the concentration of the manganese(II) ion in the anode electrolyte solution at the end of the discharge? Similarly, what will be the concentration of the copper(II) ion in the cathode electrolyte solution at the end of the discharge? Given the nature of how a cell is constructed, you would usually deplete the ions at the cathode before completely oxidizing the anode (since the latter would break the circuit). This is because the complete oxidation of the anode would likely only occur if you were using an extremely thin wire of the metal. In this case, let us assume you constructed the voltaic cell using sufficiently sized rods for each metal. This means that the total time you can run your cell will depend on the initial copper ion concentration. Calculate the maximum run time for your voltaic cell if it runs at 0.5 AA and your initial concentration of CuCl2(aq)CuCl2(aq) was 0.228 M'