1988 United States Senate election in Ohio
November 8, 1988
 
Nominee Howard Metzenbaum George Voinovich
Party Democratic Republican
Popular vote 2,480,038 1,872,716
Percentage 56.97% 42.31%
County results
Metzenbaum:      50–60%      60–70%      70–80%
Voinovich:      50–60%      60–70%
U.S. senator before election

Howard Metzenbaum
Democratic

 Elected U.S. Senator

Howard Metzenbaum
Democratic

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. In addition, Metzenbaum's 56.97% of the vote is to date the best performance for a Democrat running for the U.S. Senate in Ohio, though Sherrod Brown, a Democrat nearly broke that in 2006. Voinovich would later be elected in the other Senate seat ten years later. As of 2023, this remains the last time that Ohio would support different parties in concurrent presidential and Senate elections.

Major candidates

Democratic

Democratic primary results
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%

Republican

Republican primary results
Party Candidate Votes %
Republican George Voinovich 636,806 100.00%
Total votes 636,806 100.00%

Results

1988 United States Senate election in Ohio
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

See also

References

  1. ↑ "Our Campaigns - OH US Senate Race - Nov 08, 1988".
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