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Pyruvate is the result of glycolysis, and it has two possible outcomes. It passes via aerobic respiration in the aerobic process and fermentation in the anaerobic process. Pyruvate fermentation can take two forms: lactic acid fermentation and alcoholic fermentation.

The two reactions are given as,

笔测谤耻惫补迟别+2贬鈦�+2别鈦烩啋尝补肠迟补迟别

狈础顿鈦�+2贬鈦�+2别鈦烩啋狈础顿贬+贬鈦�

The coupled reaction is,

笔测谤耻惫补迟别+2贬鈦�+2别鈦烩啋尝补肠迟补迟别+NAD鈦�

The free energy change equation is,

$\mathrm{螖蔚}=\mathrm{螖蔚}掳\text{'}-\frac{\mathrm{RT}}{\mathrm{nF}}\ln \left(\frac{\left[\mathrm{lactate}\right]\left[{\mathrm{NAD}}^{+}\right]}{\left[\mathrm{pyruvate}\right]\left[\mathrm{NADH}\right]}\right)$

On substituting,

$\begin{array}{c}\mathrm{螖蔚}=0.130\mathrm{V}-\frac{\mathrm{RT}}{\mathrm{nF}}\ln \left(1\right)\\ 鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�=0.130\mathrm{V}\\ \mathrm{螖骋}=-\left(2\right)\left(96485{\mathrm{JV}}^{-1}{\mathrm{mol}}^{-1}\right)\left(0.130\mathrm{V}\right)\\ 鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�=-25.1{\mathrm{kJmol}}^{-1}\end{array}$

The free energy change equation is,

$\mathrm{螖蔚}=\mathrm{螖蔚}掳\text{'}-\frac{\mathrm{RT}}{\mathrm{nF}}\ln \left(\frac{\left[\mathrm{lactate}\right]\left[{\mathrm{NAD}}^{+}\right]}{\left[\mathrm{pyruvate}\right]\left[\mathrm{NADH}\right]}\right)$

On substituting,

$\begin{array}{c}\mathrm{螖蔚}=0.130\mathrm{V}-\frac{\mathrm{RT}}{\mathrm{nF}}\ln \left(160脳160\right)\\ 鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�=0\mathrm{V}\\ \mathrm{螖骋}=0\end{array}$

The free energy change equation is,

$\mathrm{螖蔚}=\mathrm{螖蔚}掳\text{'}-\frac{\mathrm{RT}}{\mathrm{nF}}\ln \left(\frac{\left[\mathrm{lactate}\right]\left[{\mathrm{NAD}}^{+}\right]}{\left[\mathrm{pyruvate}\right]\left[\mathrm{NADH}\right]}\right)$

On substituting,

$\begin{array}{c}\mathrm{螖蔚}=0.130\mathrm{V}-\frac{\mathrm{RT}}{\mathrm{nF}}\ln \left(1000脳1000\right)\\ 鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�=-0.047\mathrm{V}\\ \mathrm{螖骋}=-\left(2\right)\left(96485{\mathrm{JV}}^{-1}{\mathrm{mol}}^{-1}\right)\left(-0.047\mathrm{V}\right)\\ 鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�=9.1{\mathrm{kJmol}}^{-1}\end{array}$

The 鈭咷 is negative in part a, so the reaction is still moving ahead. Because the 鈭咷 value is zero, the part b area represents the equation in the equilibrium state. Parts c has a good positive value for 鈭咷. As a result, the reaction proceeds in a retrograde direction of its own accord.