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A voltaic cell employs the following redox reaction:
Sn^2+(aq)+ Mn(s)--> Sn(s) + Mn^2+(aq)
Calculate the cell potential at 25 under each of the following conditions.

a.)[Sn]= 1.51E-2 M [Mn]= 2.52M
b.)[Sn]= 2.52 [Mn]= 1.51E-2 M

  • chemistry - ,

    That is 25 degree celsius

  • chemistry - ,

    Obviously [Sn] can't be anything other than 1.00. You must mean [Sn^+2]. Same for Mn.
    The easiest way to handle this is to calculate the half cell potential for each half cell, then add them together.
    The reduction potential equation is
    Ehalfcell = Eo-(0.0592/n)*log(red/ox)
    For Sn that is
    Ehalfcell = Eo (look up the potential as a reduction potential), then - (0.0592/2)*log(1/0.0151) = ?? (Note: The 1.00 is the value for Sn solid and 0.0151 is the value for Sn^+2 from the problem.)

    Do the same thing for the Mn couple (but do it as a reduction), then reverse the equation, change the sign of the potential, and add the oxidation half to the reduction half. Post your work if you get stuck.

  • chemistry - ,

    Thank you

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