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Calculate the equilibrium constant for the following reaction: `3Sn + 2Cr_(2)O_(7)^(2-) + 28H^(-) rarr 3Sn^(4+) + 4Cr^(3+) + 14H_(2)O` `E_(Sn//Sn^(4+))^(@)= 0.009 V, " " E_(Cr_(2)O_(7)^(2-)//Cr^(3))^(@) = 1.33V`A. `10^(413)`B. `10^(1.8)`C. `10^(272)`D. `10^(317)` |
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Answer» Correct Answer - C Number of electrons exchanged `= 12` for 3Sn) `E_(cell)^(@) = 0.009 + 1.33 = 1.339 = (0.0591)/(12)"log" K_(eq).` `K_(eq) = 10^(271.878)` `= 10^(272)` |
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