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Here in the reaction there is an exchange of 2 electronsTherefore.pdf

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Here in the reaction there is an exchange of 2 electrons Therefore E = E0 - (0.059/2)log((Mn2+)/(H+)^2) 0.511 = 0 + 0.0295*log([H+]^2/[Mn2+]) 17.322 = log([H+]^2/1 Therfore [H+] = 4.58*10^8 pH = -log[H+] = 8.66 Solution Here in the reaction there is an exchange of 2 electrons Therefore E = E0 - (0.059/2)log((Mn2+)/(H+)^2) 0.511 = 0 + 0.0295*log([H+]^2/[Mn2+]) 17.322 = log([H+]^2/1 Therfore [H+] = 4.58*10^8 pH = -log[H+] = 8.66.

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Here in the reaction there is an exchange of 2 electrons Therefore E = E0 - (0.059/2)log((Mn2+)/(H+)^2) 0.511 = 0 + 0.0295*log([H+]^2/[Mn2+]) 17.322 = log([H+]^2/1 Therfore [H+] = 4.58*10^8 pH = -log[H+] = 8.66 Solution Here in the reaction there is an exchange of 2 electrons Therefore E = E0 - (0.059/2)log((Mn2+)/(H+)^2) 0.511 = 0 + 0.0295*log([H+]^2/[Mn2+]) 17.322 = log([H+]^2/1 Therfore [H+] = 4.58*10^8 pH = -log[H+] = 8.66

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Here in the reaction there is an exchange of 2 electronsTherefore.pdf

  • 1. Here in the reaction there is an exchange of 2 electrons Therefore E = E0 - (0.059/2)log((Mn2+)/(H+)^2) 0.511 = 0 + 0.0295*log([H+]^2/[Mn2+]) 17.322 = log([H+]^2/1 Therfore [H+] = 4.58*10^8 pH = -log[H+] = 8.66 Solution Here in the reaction there is an exchange of 2 electrons Therefore E = E0 - (0.059/2)log((Mn2+)/(H+)^2) 0.511 = 0 + 0.0295*log([H+]^2/[Mn2+]) 17.322 = log([H+]^2/1 Therfore [H+] = 4.58*10^8 pH = -log[H+] = 8.66