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The 1988 United States Senate election in Ohio was held on November 8, 1988. Incumbent Democratic U.S. Senator Howard Metzenbaum won re-election.[1] Metzenbaum easily won the Democratic nomination with over 80% of the vote, while Cleveland Mayor George Voinovich was uncontested in his primary. This was the last U.S. senator to win in the Democratic party at this seat until 2006. Voinovich would later be elected in the other Senate seat ten years later. As of 2023[update], this remains the last time that Ohio would support different parties in concurrent presidential and Senate elections.
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Metzenbaum: 50–60% 60–70% 70–80% Voinovich: 50–60% 60–70% | |||||||||||||||||
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| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Democratic | Howard Metzenbaum | 1,070,934 | 83.57% | |
| Democratic | Ralph Applegate | 210,508 | 16.43% | |
| Total votes | 1,281,442 | 100.00% | ||
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | George Voinovich | 636,806 | 100.00% | |
| Total votes | 636,806 | 100.00% | ||
| Party | Candidate | Votes | % | ||
|---|---|---|---|---|---|
| Democratic | Howard Metzenbaum (incumbent) | 2,480,038 | 56.97% | ||
| Republican | George Voinovich | 1,872,716 | 42.31% | ||
| Independent | David Marshall | 151 | 0.00% | ||
| Majority | 607,322 | 14.66% | |||
| Turnout | 4,352,905 | 100.00% | |||
| Democratic hold | |||||
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