2MnO4-(aq) + 16H+(aq) + 5Hg(l) <--> 2Mn2+ +8H2O(l) + 5Hg2+(aq) Eo = 1.51-0.854 = 0.656 V 10 electrons are transfered in the above reaction, n = 10 E = Eo - (RT/nF) ln(K) 0.620 = 0.656 - 0.0592/10 log(K) 0.036 = 0.0592/10 log(0.2002x0.1505/0.2002x[H+]16) 1.205 x 106 = 0.2002x0.1505/0.2002x[H+]16 [H+] = 0.23 M pH = - log[H+] = 0.64 Solution 2MnO4-(aq) + 16H+(aq) + 5Hg(l) <--> 2Mn2+ +8H2O(l) + 5Hg2+(aq) Eo = 1.51-0.854 = 0.656 V 10 electrons are transfered in the above reaction, n = 10 E = Eo - (RT/nF) ln(K) 0.620 = 0.656 - 0.0592/10 log(K) 0.036 = 0.0592/10 log(0.2002x0.1505/0.2002x[H+]16) 1.205 x 106 = 0.2002x0.1505/0.2002x[H+]16 [H+] = 0.23 M pH = - log[H+] = 0.64.