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Unformatted text preview: estrada (gse77) – Homework 11 – Sutcliffe – (52410) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. A NOTE ON CELL DIAGRAMS: There is no good explanation for this but when the diagram is more complex than A  B  C  D then you should just assume everything on the LEFT of the  is for the ANODIC reac tion, and everthing on the RIGHT of the  is the CATHODIC reaction. Then for each of these, find a reduction half reaction in ta ble 1.1 that includes all of the items in your anodic or cathodic reaction, and use that to write down your two half reactions. Unless Pt is listed on the far left or right; in that case it is just an inert electrode, ignore it. If you look at the given half reactions for Q2 (es pecially the cathodic reaction), youll see how this works. 001 10.0 points Consider the cell Zn(s)  Zn 2+ (aq , . 100 M)  Cl − (aq , ? M)  Cl 2 (g , . 500 atm)  Pt . For this cell, E ◦ = 2.12 V and E = 2.27 V at 25 ◦ C. Calculate the Cl − (aq) concentration in the cathode compartment. 1. 1.2 × 10 − 1 M 2. 1.5 × 10 − 3 M 3. 2.9 × 10 − 3 M 4. 6.5 × 10 − 3 M correct 5. 4.3 × 10 − 5 M Explanation: 002 10.0 points Consider the cell Pt  H 2 (1 atm); H + (? M)  Hg 2 Cl 2 (s); Cl − (1 M)  Hg 2 H + + 2 e − → H 2 E = 0 . 00 V Hg 2 Cl 2 + 2 e − → 2 Hg + 2 Cl − E = 0 . 268 V If the measured cell potential for the cell is . 35 volts, what is the pH of the solution? 1. 5.45 2. 4.74 3. 2.77 4. less than 1.00 5. 1.39 correct Explanation: [H + ] is not known, likely not at 1 M (stan dard conditions). The second reaction is un der standard conditions. So we know that: H 2 → 2 H + + 2 e − E = ? 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 . 0592 n log parenleftbigg [Red] y [Ox] x parenrightbigg . 082 V = 0 V . 0592 2 log parenleftbigg 1 x 2 parenrightbigg . 082 V = . 0592 2 log(1) + . 0592 2 log x 2 . 082 V = 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 003 10.0 points What is the standard free energy change for the reaction H 2 + 2 Ag + (aq) → 2 H + (aq) + 2 Ag(s)? The standard potentials are 2 H + + 2 e − → H 2 (0 V) Ag + + e − → Ag (0 . 7994 V) Correct answer: 154 . 261 kJ. estrada (gse77) – Homework 11 – Sutcliffe – (52410) 2 Explanation: 004 10.0 points Consider two hypothetical metals, X and Y, which can exist as metal ions X 2+ and Y 3+ , respectively. For these two metals, the standard potentials are X 2+ + 2 e − → X(s) ( . 1 V) Y 3+ + 3 e − → Y(s) (0 . 1 V) Calculate Δ G for the reaction 3 X(s) + 2 Y 3+ (0 . 2251 M) → 3 X 2+ (0 . 2282 M) + 2 Y(s) ....
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 Spring '07
 Holcombe
 Electrochemistry, Reaction, Chemical reaction, Rate equation, ESTRADA

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