| Election Name: | 1988 United States Senate election in Ohio |
| Country: | Ohio |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1982 United States Senate election in Ohio |
| Previous Year: | 1982 |
| Next Election: | 1994 United States Senate election in Ohio |
| Next Year: | 1994 |
| Election Date: | November 8, 1988 |
| Nominee1: | Howard Metzenbaum |
| Party1: | Democratic Party (United States) |
| Popular Vote1: | 2,480,038 |
| Percentage1: | 56.97% |
| Nominee2: | George Voinovich |
| Party2: | Republican Party (United States) |
| Popular Vote2: | 1,872,716 |
| Percentage2: | 42.31% |
| Map Size: | 210px |
| U.S. Senator | |
| Before Election: | Howard Metzenbaum |
| Before Party: | Democratic Party (United States) |
| After Election: | Howard Metzenbaum |
| After Party: | Democratic Party (United States) |
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., this remains the last time that Ohio would support different parties in concurrent presidential and Senate elections.