3.) Voltaic cell below with resistance of 3.13 ohm. Current flow through a solution is 30.0 mA. Solve for the voltage (i

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3.) Voltaic cell below with resistance of 3.13 ohm. Current flow through a solution is 30.0 mA. Solve for the voltage (i

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3.) Voltaic cell below with resistance of 3.13 ohm. Current flowthrough a solution is 30.0 mA. Solve for the voltage (in volts)applied to drive the reaction.
Hg(l)|Hg2Cl2(s)|KCl(saturated)||KCl(0.90 M) |Cl2 (grams,0.29 atm)|Pt(s)
Anode : Hg2Cl2(s) + 2e- 2Hg(l) + 2Cl- E0 = 1.29 V
Cathode : Cl2(g) + 2e-2Cl- E0=2.80 V
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