Copper reduces $$NO3$$− into NO and $$NO2$$ depending upon the concentration of $$HNO3$$ in solution [Assuming fixed $$Cu2+$$ $$andPNO=PNO2$$ $$)$$ , the concentration at which the thermodynamic tendency for reduction of $$NO3$$− into NO and $$NO2$$ by copper is same is $$10xM$$. The value of $$2x$$ is……. (Rounded$$ off to the nearest $$integer) [Given $$ECu2+/Cu$$°$$=0.34V$$, $$ENO3$$−$$/NO$$°$$=0.96V$$,$$ENO3$$−$$/NO2$$°$$=0.79V$$ and $$298K$$ $$RTF(2.303)=0.059$$]

Question

Copper reduces $$NO3$$− into NO and $$NO2$$ depending upon the concentration of  $$HNO3$$ in solution [Assuming fixed $$Cu2+$$ $$andPNO=PNO2$$ $$)$$ , the concentration at which the thermodynamic tendency for reduction of $$NO3$$−  into NO and $$NO2$$ by copper is same is $$10xM$$. The value of $$2x$$ is……. (Rounded$$  off to the nearest $$integer) [Given $$ECu2+/Cu$$°$$=0.34V$$,  $$ENO3$$−$$/NO$$°$$=0.96V$$,$$ENO3$$−$$/NO2$$°$$=0.79V$$ and $$298K$$ $$RTF(2.303)=0.059$$]

Infinity Learn · Accepted Answer

$$3Cu+8H++2NO3-$$→$$3Cu+2+2NO+4H2OCu+4H++2NO3-$$→$$Cu2++2NO2+2H2O$$ Let con of $$HNO3$$ $$=H+=x$$   $$NO3$$−$$=x$$ $$ENO3$$−$$/NO$$ −$$ECu2+/Cu=ENO3$$−$$/NO2$$−−$$ECu2+/Cu$$                       $$ENO3$$−$$/NO=ENO3$$−$$/NO2$$−  $$0.96$$−$$0.05913log10$$−$$3105=0.79$$ −$$0.0593log$$ $$10$$−$$3x3$$        $$X=0.66$$      $$2x=1.32$$  nearst integer is $$1$$

Detailed Solution

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