Question

. Your calculations on the first page centered around: Δe--nFE cell, Where: AG describes the free enera vailable from a system, in units of n describes the os e Fis Faraday's constant, the charge, measured in CO E cll is Polenheal which can also be written asJgule per permo transferred in a chemical reaction f electrons. er om 2. On the first week of lab, the metals that reacted most vigorously were silver and magnesium In your well trays, we saw the reaction: a. Balance this equation. What is n for this reaction as written? b. If this cell has a voltage in standard conditions (Eocell) of 3.17 V, what is ΔG? (in kJ/mol) c. If you used 20 mg of silver in this reaction (about what you would have had in your solution in lab), but turned it in to a battery, how much energy could you get out of that battery? d. How many Coulombs of charge would be moved as you discharged the battery? e. Let's suppose you discharged the battery over the space of 5 minutes. What would the current be (in Amps)

Answer 1

When Silver and Magnesium reacts vigorously following is the equation

(i)

Mg(s) → Mg2+ (aq) + 2e-

(ii)

Ag+ (aq) + e-→ Ag(s)

By multiplying reaction (ii) with 2 and from combining the above two reactions, we get

Mg (s) + 2Ag+ (aq) → 2Ag (s) Mg2+ (aq)

(c) If 20mg of silver is used than the energy we get from that battery is

∆G = -nFE⁰_cell

Where n = number of moles of electron = weight of silver in gram/ molecular weight of silver

n = 0.02 g/ 107.8682 = 1.85 × 10^-4

F = 96485 C/mol

E⁰_cell = 3.17 V

Therefore, ∆G = -nFE⁰_cell

                     = -(1.85 × 10^-4 ) (96485 C/mol) (3.17 V)

                     = - 56.58 KJ/mol

(d) When battery is discharged coloumbs of charge is

Q = number of electron × Faraday constant

    = 2 × 96485 C = 192970 Coloumbs

(e) Q = I × t

Where, I = current in Amp; t = time in seconds

I = Q/t = 192970 C / 300 sec = 643.23 Amp