# Hg 2 cl 2 2 e 2 hg 2 cl e 0 268 v e cell 0 35 v e for

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Hg 2 Cl 2 + 2 e 2 Hg + 2 Cl E = +0 . 268 V E cell = +0 . 35 V E for the first reaction must be 0 . 082 V. Use the Nernst equation to determine [H + ]: E = E 0 - 0 . 0592 n log parenleftbigg [Red] y [Ox] x parenrightbigg - 0 . 082 V = 0 V - 0 . 0592 2 log parenleftbigg 1 x 2 parenrightbigg - 0 . 082 V = - 0 . 0592 2 log(1) + 0 . 0592 2 log x 2 - 0 . 082 V = 0 + 0 . 0592 2 log x 2 - 2 . 77027 = log x 2 x 2 = 10 2 . 77027 x = 0 . 0411969 = [H + ] pH = - log[H + ] = - log(0 . 0411969) = 1 . 38514

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Homework #10 – holcombe – 3 008 10.0 points A battery formed from the two half re- actions below dies (reaches equilibrium). If [Fe 2+ ] was 0 . 24 M in the dead battery, what would [Cd 2+ ] be in the dead battery? Half reaction E Fe 2+ -→ Fe - 0 . 44 Cd 2+ -→ Cd - 0 . 40 1. 5 . 4 M 2. 120 . 3 M 3. 0 . 01 M correct 4. 0 . 0 . 0005 M Explanation: E = +0 . 04 E cell = E cell - 0 . 05916 N e log Q 0 = 0 . 04 - 0 . 05916 2 log 0 . 24 [Cd 2+ ] log 0 . 24 [Cd 2+ ] = 1 . 35 0 . 24 [Cd 2+ ] = 10 1 . 35 [Cd 2+ ] = 0 . 24 10 1 . 35 = 0 . 0107 M 009 10.0 points Consider the voltaic cell In | In 3+ || Ru 3+ , Ru 2+ | Pt In 3+ + 3 e In E 0 = - 0 . 340 V Ru 3+ + 1 e Ru 2+ E 0 = - 0 . 080 V at 25 C. What is the equilibrium constant for the overall cell reaction? 1. 1 . 3 × 10 1 2. 7 . 1 × 10 17 3. 2 . 4 × 10 4 4. 1 . 5 × 10 13 correct 5. 1 . 2 × 10 7 6. 2 . 0 × 10 21 Explanation: 010 10.0 points The standard voltage of the cell Pt | H 2 (g) | H + (aq) || Cl (aq) | AgCl(s) | Ag(s) is +0.22 V at 25 C. Calculate the equilibrium constant for the reaction 2 AgCl(s) + H 2 (g) 2 Ag(s) + 2 H + (aq) + 2 Cl (aq) 1. 7.4 2. 2 . 7 × 10 7 correct 3. 5 . 2 × 10 3 4. 3.7 5. 1 . 7 × 10 3 Explanation: 011 10.0 points The principle of inhibiting the corrosion of iron by using a sacrificial anode is to allow 1.
• Fall '07
• Holcombe

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