Answer
$E_{cell} = 0.83 V$
Work Step by Step
$2H^{+}_{(aq)} + Pb_{(s)} → H_{2(g)} + Pb^{2+}_{(aq)}$
Here n= 2, since the equation involves 2 electrons for charge balancing.
$E_{cell} = E^{0}_{cell} – (RT \div 2F)ln([Pb^{2+}][H_{2}] \div [H^{+}]^{2})$
$E^{0}_{cell} = E^{0}_{H^{+}/H_{2}} - E^{0}_{Pb^{2+}/Pb}$
$E^{0}_{cell} = 0.00-(-0.13) = 0.13 V$
$[Pb^{2+}] = 0.10 M$ ,$ [H^{+}] = 0.050 M$ $P_{H_{2}}= 1 atm $
$E_{cell} = 0.13 V – (RT \div 2F)ln(0.10 \times 1 \div [0.050]^{2})$
$E_{cell} = 0.13 V – (2.303\times8.314 J K^{-1}\times 298) \div (2\times96500)log(0.10 \div 0.0025)$
$E_{cell} = 0.13 V –0.0296 log(40))$
$E_{cell} = 0.13 V –0.0296 \times 1.602$
$E_{cell} = 0.13 V –0.047= 0.83 V$
