The equilibrium constant for the following reaction at `25^(@)C " is " 2.9 xx 10^(9)`. Calculate standard voltage of the cell.
`C1_(2(g)) + 2Br_((aq))^(-) hArr Br_(2(1)) + 2C1_((aq))^(-)`

Answer

Standard voltage of the cell `=E_("cell ")^(0) = 0.28 V`

Explanation

Given : Cell reaction `:C1_(2(g)) + 2Br_((aq))^(-) hArr Br_(2(1)) + 2C1_((aq))^(-)`
Equilibrium constant = k = 2.9 `xx10^(-9) atm ^(-1)`
Standard voltage of the cell = `E_("cell ")^(0) = ? `
The formulation of the cell :
`Pt|Br_(2(1)) | (1M)" ||"C1_((aq))^(-) (1M)|C1_(2) (g,P_(C1_(2))) | Pt`
`((LHE 2Br_((aq))^(-) to Br_(2(1)) + 2e^(-))/(RHE . C1_(2(g)) + 2e^(-) to 2C1_((aq))^(-)))/(2Br_((aq))^(-) + C1_(2(g)) to Br_(2(1))+ 2C1_((aq))^(-))" "underset"(Overall cell reaction )"underset"(Reduction at cathode )"("(Oxidation at anode)")`
`:. n=2`
`E_("cell ")^(0) = (0.0592)/(n) log_(10) K`
`= (0.0592)/(2) log_(10) 2.9 xx 10^(9)`
`=0. 0296 xx (9.4624)`
`=0.28 V`