Redox and Electrochemistry Practice Questions & Answers

According to the mnemonic OIL RIG (Oxidation Is Loss, Reduction Is Gain), oxidation is defined as the loss of electrons.

The reducing agent loses electrons (is oxidized) and thereby causes the reduction of another species. Therefore, the electron donor is the reducing agent.

$Cu^{2+}$Cu2+ gains electrons to become $Cu(s)$Cu(s). Since it accepts electrons, it causes the oxidation of zinc, making $Cu^{2+}$Cu2+ the oxidizing agent.

Sodium ($Na$Na) starts with an oxidation number of 0 and ends with +1 in $NaCl$NaCl. An increase in oxidation number indicates oxidation.

Reduction is the gain of electrons. In option b, the chlorine molecule gains two electrons to form chloride ions ($Cl^-$Cl−).

In $H_2SO_4$H2​SO4​: H is +1 (total +2), O is -2 (total -8). Let S be $x​$x​.
$2(+1) + x + 4(-2) = 0$2(+1)+x+4(−2)=0
$2 + x - 8 = 0 \Rightarrow x = +6$2+x−8=0⇒x=+6

In $MnO_4^-$MnO4−​: O is -2 (total -8). The overall charge is -1.
$x + 4(-2) = -1$x+4(−2)=−1
$x - 8 = -1 \Rightarrow x = +7$x−8=−1⇒x=+7

Peroxides are a major exception to the oxygen rule. In $H_2O_2$H2​O2​, H is +1, so:
$2(+1) + 2x = 0 \Rightarrow 2x = -2 \Rightarrow x = -1$2(+1)+2x=0⇒2x=−2⇒x=−1

In $Cr_2O_7^{2-}$Cr2​O72−​: O is -2 (total -14). Overall charge is -2.
$2x + 7(-2) = -2$2x+7(−2)=−2
$2x - 14 = -2 \Rightarrow 2x = +12 \Rightarrow x = +6$2x−14=−2⇒2x=+12⇒x=+6

In $NH_4^+$NH4+​: H is +1 (total +4). Overall charge is +1.
$x + 4(+1) = +1$x+4(+1)=+1
$x + 4 = +1 \Rightarrow x = -3$x+4=+1⇒x=−3