Samuel Dominic Chukwuemeka (SamDom For Peace)

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Solved Examples on Redox Reactions

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For Questions (1.) through (12.); calculate the: oxidation number, and give the IUPAC name of the substance.

(1.) $N$ in $N_2O$


Let the oxidation number of nitrogen = $x$

$ N = 2 * x = 2x \\[3ex] O = 1 * -2 = -2 \\[3ex] N_2O = 0 \\[3ex] 2x - 2 = 0 \\[3ex] 2x = 2 \\[3ex] x = +1 \\[3ex] $ The IUPAC name of $N_2O$ is Dinitrogen(I)oxide
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(2.) $N$ in $NO$


Let the oxidation number of nitrogen = $x$

$ N = 1 * x = x \\[3ex] O = 1 * -2 = -2 \\[3ex] NO = 0 \\[3ex] x - 2 = 0 \\[3ex] x = +2 \\[3ex] $ The IUPAC name of $NO$ is Nitrogen(II)oxide
(3.) $N$ in $NO_2$


Let the oxidation number of nitrogen = $x$

$ N = 1 * x = x \\[3ex] O = 2 * -2 = -4 \\[3ex] NO_2 = 0 \\[3ex] x - 4 = 0 \\[3ex] x = +4 \\[3ex] $ The IUPAC name of $NO_2$ is Nitrogen(IV)oxide
(4.) $N$ in $N_2O_5$


Let the oxidation number of nitrogen = $x$

$ N = 2 * x = 2x \\[3ex] O = 2 * -5 = -10 \\[3ex] N_2O_5 = 0 \\[3ex] 2x - 10 = 0 \\[3ex] 2x = 10 \\[3ex] x = +5 \\[3ex] $ The IUPAC name of $N_2O_5$ is Dinitrogen(V)oxide
(5.) $Cr$ in $K_2Cr_2O_7$


Let the oxidation number of chromium = $x$

$ K = 2 * 1 = 2 \\[3ex] Cr = 2 * x = 2x \\[3ex] O = 7 * -2 = -14 \\[3ex] K_2Cr_2O_7 = 0 \\[3ex] \therefore 2 + 2x - 14 = 0 \\[3ex] 2x - 12 = 0 \\[3ex] 2x = 12 \\[3ex] x = +6 \\[3ex] $ The IUPAC name of $K_2Cr_2O_7$ is Potassium heptaoxodichromate(VI)
(6.) $N$ in $Zn(NO_3)_2$


Let the oxidation number of nitrogen = $x$

$Zn = 1 * 2 = 2$ Electronic Configuration of $Zn = 30 = 2, 8, 18, *2*$

$ N = 2 * x = 2x \\[3ex] O = 2 * 3 * -2 = -12 \\[3ex] Zn(NO_3)_2 = 0 \\[3ex] \therefore 2 + 2x - 12 = 0 \\[3ex] 2x - 10 = 0 \\[3ex] 2x = 10 \\[3ex] x = +5 \\[3ex] $ The IUPAC name of $Zn(NO_3)_2$ is Zinc trioxonitrate(V)
(7.) $B$ in $NaBH_4$


Let the oxidation number of boron = $x$

$ Na = 1 * 1 = 1 \\[3ex] B = 1 * x = x \\[3ex] H = 4 * -1 = -4 \:\:Na\:\: and\:\: B \:\:are\:\: metals... Rule\:\: 9 \\[3ex] NaBH_4 = 0 \\[3ex] \therefore 1 + x - 4 = 0 \\[3ex] x - 3 = 0 \\[3ex] x = +3 \\[3ex] $ The IUPAC name of $NaBH_4$ is Sodium tetrahydroborate(III)
(8.) ACT For all real values of $x$, which of the following equations is true?

$ F.\:\: \sin(7x) + \cos(7x) = 7 \\[3ex] G.\:\: \sin(7x) + \cos(7x) = 1 \\[3ex] H.\:\: 7\sin(7x) + 7\cos(7x) = 14 \\[3ex] J.\:\: \sin^2(7x) + \cos^2(7x) = 7 \\[3ex] K.\:\: \sin^2(7x) + \cos^2(7x) = 1 $


$ F.\:\: -\sin(-x^\circ) = -1 * \sin(-x) \\[3ex] \sin(-x) = -\sin x ...Odd\:\: Identity \\[3ex] = -1 * -\sin x = \sin x ...YES \\[5ex] G.\:\: \sin(-x^\circ) \\[3ex] \sin(-x) = -\sin x ...Odd\:\: Identity \\[3ex] -\sin x \ne \sin x ...NO \\[5ex] H.\:\: \cos(90 - x)^\circ \\[3ex] \cos(90 - x) = \sin x ...Cofunction\:\: Identity \\[3ex] YES \\[5ex] J.\:\: \cos(x - 90)^\circ \\[3ex] -(x - 90) = -x + 90 = 90 - x \\[3ex] \cos(x) = \cos(-x) ...Even\:\: Identity \\[3ex] \cos(x - 90) = \cos(90 - x) ...Even\:\: Identity \\[3ex] = \sin x ...YES \\[5ex] K.\:\: \sqrt{1 - (\cos x^\circ)^2} \\[3ex] (\cos x)^2 = \cos^2 x \\[3ex] 1 - \cos^2 x = \sin^2 x ...Pythagorean\:\: Identity \\[3ex] = \sqrt{\sin^2 x} = \sin x ...YES $